How much money will Rachel have in her account in ten years
TRUE OR FALSE! If a quadratic equation can be factored and each factor contains only real numbers then there can not be an imaginary solution.
3 is 1 percent of what amount
The value of the solution is, 3 is 1 percent of 300.
We have to give that,
To find the amount for 1 percent is 3.
Let us assume that,
3 is 1 percent of x.
Hence, It can be written as,
3 = 1% of x
Solve for x,
3 = 1/100 × x
3 × 100 = x
x = 300
Therefore, 3 is 1 percent of 300.
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a girl has a coloring book with 60 pages in it. she uses 5 pages every 3 days. How many days will it take her to finish the coloring book?
Corresponding angles lie on the same side of a transversal?
SELECT ONE:
TRUE
FALSE
Forty-six and seven thousandths in decimal form
Find the area of a parallelogram with sides of 12 inches and 8 inches if one of the angles is 120
degrees
Answer:
48√3 sq. in.
Step-by-step explanation:
I know this is correct bc I just had this question and this was the correct answer
Five cards are randomly selected without replacement from a standard deck of 52 playing cards. what is the probability of getting 5 hearts? round your answer to four decimal places
The probability of getting 5 hearts from the deck of cards is 0.0962.
We need to find the probability of getting 5 hearts.
What is in a 52 deck of cards?A "standard" deck of playing cards consists of 52 Cards in each of the 4 suits of Spades, Hearts, Diamonds, and Clubs. Each suite contains 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.
P(E)=Number of favourable outcomes/Total number of outcomes.
Now, P(E)=5/52=0.0962
Therefore, the probability of getting 5 hearts from the deck of cards is 0.0962.
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The height of a coconut falling from a tree can be represented by the function h(t)=-16t^2 + 24, where h(t) is the height of the coconut, in feet, and t is time, in seconds.
What is the initial height, in feet, of the coconut?
Answer:
The answer is C "The values of h(t) when t = 4 and 5 should be 0."
Ava was making muffins. she used 1 1/3 tsp of cinnamon and 1/2 tsp of nutmeg. how many teaspoons of spice did ava use?
The range of the function f(x)=x+5 is (7,9). What is the function's domain?
Is it
(2,4)
(-2,-4)
(12,14)
(-12,-14)
(0,5)
When it is 2 hours after 2 o'clock, then it is 4 o'clock (2 + 2 = 4). When it is 10 hours after 10 o'clock, then it is 8 o'clock. In this kind of "clock arithmetic," 10 + 10 = 8.
When a clock time gets bigger than 12, you subtract 12 and take the answer as the actual clock time. For example, if you subtract 12 from 20, the answer is 8, so 20 o'clock is really 8 o'clock.
Brad has a certain medication that he needs to take every 5 hours without fail, starting at 1 o'clock on a certain day. The sequence of clock times that he takes his pills is 1, 6, 11, 4, 9, ...
What is the clock time when Brad takes his 16th pill?
Stella graphs the equation y=13x – 2y=13x – 2 .
Select all statements about Stella's graph that are true.
The graph is a straight line.
The line passes through the origin.
The line passes through the point (0, –2)(0, –2) .
The slope of the line is 3.
The y-intercept of the line is 2.
Given that point U is the circumcenter of triangle XVZ, which segments are congruent?
Answer: [tex]\overline{WX}\cong \overline{WV},\ \overline{VA}\cong\overline{AZ}[/tex]
[tex]\overline{XY}\cong\overline{YZ}[/tex]
[tex]\overline{UV}\cong \overline{UZ}\cong \overline{UX}[/tex]
Step-by-step explanation:
In the given figure we have a triangle , in which U is the circumcenter of triangle XVZ.
We know that the circumcenter is equidistant from each vertex of the triangle.
Since , the line segments which are representing the distance from the vertex and the circumcenter are [tex]\overline{UV},\ \overline{UZ},\ \overline{UX}[/tex]
Also, The circumcenter is at the intersection of the perpendicular bisectors of the triangle's sides.
Then , [tex]\overline{WX}\cong \overline{WV},\ \overline{VA}\cong\overline{AZ}[/tex] and [tex]\overline{XY}\cong\overline{YZ}[/tex]
Hence, the segments which are congruent are [tex]\overline{UV}\cong \overline{UZ}\cong \overline{UX}[/tex]
[tex]\overline{WX}\cong \overline{WV},\ \overline{VA}\cong\overline{AZ}[/tex]
[tex]\overline{XY}\cong\overline{YZ}[/tex]
what is f (x)=2x^2+5 multiplied by g (x)=2x
A) 7x+2x
B) 4x+10x
C)4x^2+10x
D)4x^3+10x
what is 26.100??
...?
{4x+3y=6
{2x -5y=16
Which of the following points is the solution to the system?
cos2x- sqrt 2 sinx=1 Find all solutions
To solve the equation cos2x - sqrt 2 sinx = 1, rewrite cos2x as 2cos^2x - 1. Use the quadratic formula to solve for cosx. Substitute the values back into the equation cos2x - sqrt 2 sinx = 1 and solve for x.
Explanation:To solve the equation cos2x - √2 sinx = 1, we can use trigonometric identities and equations. First, we can rewrite cos2x as 2cos^2x - 1. So, the equation becomes 2cos^2x - √2 sinx - 1 = 0. To solve this quadratic equation, let's set 2cos^2x - √2 sinx - 1 = 0 and solve for cosx.
Next, we can use the quadratic formula to solve for cosx. The quadratic formula states that x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 2, b = -√2 sinx, and c = -1. Plugging in these values, we can solve for cosx.
After solving for cosx, we can substitute the values back into the equation cos2x - √2 sinx = 1 and solve for x. The student's question is about solving the equation cos(2x) - √2 sin(x) = 1 for all solutions. We can use the trigonometric identities cos(2x) = 1 - 2sin2(x) or cos(2x) = 2cos2(x) - 1 to rewrite the equation. Since we have a sine term in the original equation, let's use the former identity:
cos(2x) - √2 sin(x) = 1
(1 - 2sin2(x)) - √2 sin(x) = 1
2sin2(x) + √2 sin(x) - 1 = 0
This is a quadratic equation in sin(x).
We can solve this quadratic equation for sin(x), then find x using inverse trigonometric functions. By factoring the quadratic or using the quadratic formula, we get solutions for sin(x). Then, we solve for x by considering all possible angles in the unit circle that correspond to the found sine values.
The equation cos(2x) - √2 sin(x) = 1 can be solved by using trigonometric identities to simplify and factor the equation, leading to solving for sin(x) using inverse operations.
Explanation:The original equation given is cos(2x) - √2 sin(x) = 1. To solve this equation, we can use trigonometric identities to simplify the cosine term. One such identity is cos(2x) = 1 - 2sin²(x), which allows us to rewrite the equation as 1 - 2sin²(x) - √2 sin(x) = 1. From there, we subtract 1 from both sides, thus isolating the sine terms on the left: - 2sin²(x) - √2 sin(x) = 0. Factoring out the common term sin(x), we get sin(x)(-2sin(x) - √2) = 0. Setting each factor equal to zero gives us two possible solutions: sin(x) = 0 and sin(x) = -√2/2. In the context of a right triangle, these solutions correspond to specific angles where the sine value is 0 and -√2/2, respectively. The solutions can be found using standard trigonometric unit circle values or by calculating the inverse sine for -√2/2.
How much simple interest would you earn for 5 years at 7% with a beginning principal of $8,000.00
A. $2,800 B. $3,200 C. $3,300 D. $3,500 I couldn't figure it out and I need some help asap.
Verify the identity.
tan^5x = tan^3xsec^2x -tan^3x
What are the factors of the expression? 3⋅(4r+y) Drag the factors of the term into the box.
Use the remainder theorem to determine whether x - 2 is a factor of
f(x) = x^3 + 3x^2 - x - 18
A) Yes, x - 2 is a factor of f(x) because f(2) = 0
B) No, x - 2 is not a factor of f(x) because f(2) = 0
C) Yes, x - 2 is a factor of f(x) because f(-2) = -12
D) No, x - 2 is not a factor of f(x) because f(-2) = -12
The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?
The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is -3/4.
Explanation:The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) can be found using the concept of implicit differentiation. To find the slope, we need to differentiate the equation with respect to x and then substitute the coordinates (2, -1) into the resulting equation. Let's solve it step by step.
First, we differentiate the equation implicitly with respect to x:
2y * dy/dx + 3(xy + 1)^2 * (y + x * dy/dx) = 0
Next, we substitute the values x = 2 and y = -1 into the equation:
2(-1) * dy/dx + 3(2(-1) + 1)^2 * (-1 + 2 * dy/dx) = 0
Simplifying the equation:
-2dy/dx - 3 * 1 * (-1 + 2dy/dx) = 0
-2dy/dx + 3 + 6dy/dx = 0
Combining like terms:
4dy/dx = -3
Finally, solving for dy/dx, we get:
dy/dx = -3/4
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The slope of the tangent line to the curve at the point \((2, -1)\) is:
[tex]\[\boxed{\frac{3}{4}}\][/tex]
To find the slope of the tangent line to the curve given by the equation [tex]\( y^2 + (xy + 1)^3 = 0 \)[/tex] at the point [tex]\( (2, -1) \)[/tex], we need to use implicit differentiation.
Given the curve:
[tex]\[y^2 + (xy + 1)^3 = 0\][/tex]
We differentiate both sides with respect to x . Using the chain rule and implicit differentiation, we get:
[tex]\[\frac{d}{dx} [y^2] + \frac{d}{dx} [(xy + 1)^3] = 0\][/tex]
First, differentiate [tex]\( y^2 \):[/tex]
[tex]\[\frac{d}{dx} [y^2] = 2y \frac{dy}{dx}\][/tex]
Next, differentiate [tex]\( (xy + 1)^3 \)[/tex] using the chain rule:
[tex]\[\frac{d}{dx} [(xy + 1)^3] = 3(xy + 1)^2 \cdot \frac{d}{dx} [xy + 1]\]\[= 3(xy + 1)^2 \cdot (y + x \frac{dy}{dx})\][/tex]
Putting it all together, we get:
[tex]\[2y \frac{dy}{dx} + 3(xy + 1)^2 (y + x \frac{dy}{dx}) = 0\][/tex]
Now, substitute [tex]\( x = 2 \) and \( y = -1 \)[/tex] into the equation:
[tex]\[2(-1) \frac{dy}{dx} + 3((2)(-1) + 1)^2 \left( -1 + 2 \frac{dy}{dx} \right) = 0\]\[-2 \frac{dy}{dx} + 3(-2 + 1)^2 \left( -1 + 2 \frac{dy}{dx} \right) = 0\][/tex]
[tex]\[-2 \frac{dy}{dx} + 3(-1)^2 \left( -1 + 2 \frac{dy}{dx} \right) = 0\]\[-2 \frac{dy}{dx} + 3(1) \left( -1 + 2 \frac{dy}{dx} \right) = 0\][/tex]
[tex]\[-2 \frac{dy}{dx} + 3(-1 + 2 \frac{dy}{dx}) = 0\]\[-2 \frac{dy}{dx} + 3(-1 + 2 \frac{dy}{dx}) = 0\][/tex]
[tex]\[-2 \frac{dy}{dx} + 3(-1) + 6 \frac{dy}{dx} = 0\][/tex]
[tex]\[-2 \frac{dy}{dx} - 3 + 6 \frac{dy}{dx} = 0\][/tex]
[tex]\[4 \frac{dy}{dx} - 3 = 0\][/tex]
[tex]\[4 \frac{dy}{dx} = 3\][/tex]
[tex]\[\frac{dy}{dx} = \frac{3}{4}\][/tex]
So, the slope of the tangent line to the curve at the point \((2, -1)\) is:
[tex]\[\boxed{\frac{3}{4}}\][/tex]
Shape 1 and shape 2 are plotted on a coordinate plane. Which statement about the shapes is true?
Shape 1 is congruent to shape 2, which can be shown using a sequence of dilations and translations.
Shape 1 is not congruent to shape 2 because the shapes do not have the same absolute coordinates.
Shape 1 is congruent to shape 2, which can be shown using a translation.
Shape 1 is not congruent to shape 2 because a sequence of rigid transformations will not map shape 1 onto shape 2.
Answer:
D
Step-by-step explanation:
(Michigan online school.)
What is 666 divided by 22?
Answer:
30.27272727272727 and repeating.
Step-by-step explanation:
What is 3 6/10 simplified?
Which expression is equivalent to 7(xy)
7x+y
7x-y
x(7y)
xy/7
Answer: [tex]x(7y)[/tex]
Step-by-step explanation:
The given expression: [tex]7(xy)[/tex]
i.e. a product of 7 and xy.
The operation used here: Multiplication.
Commutative property of multiplication :-
[tex]a\times b=b\times a[/tex] for any numbers a and b.
Associative property of multiplication :-
[tex]a\times(b\times c)=(a\times b\times c)[/tex] for any numbers a , band c.
Now, [tex]7(xy)=(7x)y[/tex] [Associative property of multiplication]
[tex]=(x7)y[/tex] [Commutative property of multiplication]
[tex]=x(7y)[/tex] [Associative property of multiplication]
what is the solution to this system of equations? 5X + 2y =29 X + 4y=13
The solution of the system of equation [tex]5x+2y = 29[/tex] ; [tex]x+4y =13[/tex] will be (3,2) and this can be determined by using the arithmetic operations.
Given :
[tex]5x+2y = 29[/tex] ----- (1)
[tex]x+4y =13[/tex] ----- (2)
Now, solve for x in equation (2).
[tex]x = 13-4y[/tex] --- (3)
Now, put the value of x obtained above in equation (1).
[tex]5(13-4y)+2y=29[/tex]
[tex]65-20y+2y=29[/tex]
[tex]36=18y[/tex]
[tex]y=2[/tex]
Now, put the value of y in equation (3).
[tex]x=13-(4\times 2)[/tex]
[tex]x=5[/tex]
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Triangle ABC is similar to triangle DEF.
What is the scale factor of triangle DEF to triangle ABC?
Triangle A B C and triangle D E F are drawn. Side A B is labeled 9. Side A C is labeled 12. Side D E is labeled 3. Side D F is labeled x.
3
1/3
4
1/4
Triangle ABC is similar to triangle DEF.
What is the scale factor of triangle DEF to triangle ABC?
Triangle A B C and triangle D E F are drawn. Side A B is labeled 9. Side A C is labeled 12. Side D E is labeled 3. Side D F is labeled x.
3
1/3
4
1/4
In the diagram, P1P2 and Q1Q2are the perpendicular bisectors of AB¯¯¯¯¯ and BC¯¯¯¯¯, respectively. A1A2 and B1B2 are the angle bisectors of ∠A and ∠B, respectively. What is the center of the circumscribed circle of ΔABC?
a. p
b. q
c. r
d. s