Convert 0.79 tons to pounds?

The smallest possible length of one of the two congruent sides is 5.7 cm and this can be determined by using the Pythagorean theorem for an acute triangle.

Given :

The longest side of an acute isosceles triangle is 8 centimeters.

The following steps can be used in order to determine the smallest possible length of one of the two congruent sides:

Step 1 - The Pythagorean theorem for an acute triangle can be used in order to determine the smallest possible length of one of the two congruent sides.

Step 2 - Let the length of the smaller sides be 'a' and let the length of the larger side be 'b'.

Step 3 - According to the Pythagorean theorem for an acute triangle:

[tex]\rm H^2 < B^2+P^2[/tex]

where H is the hypotenuse, B is the base, and P is the perpendicular.

Step 4 - Substitute the values of the known terms in the above expression.

[tex]a^2+a^2>b^2[/tex]

[tex]2a^2>b^2[/tex]

Step 5 - Substitute the value of x in the above expression.

[tex]2a^2>8^2[/tex]

a > [tex]\sqrt{32}[/tex]

a > 5.6

Therefore, the correct option is D).

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The statement is false as the Student's t-distribution is actually more peaked and has thicker tails with smaller sample sizes, and it becomes similar to the standard normal distribution as the sample size, and consequently the degrees of freedom, increase.

The statement that every Student's t-distribution with n < N, where n is the sample size and N is the population size, will be less peaked and have thinner tails is false. In reality, the t-distribution is more peaked and has thicker tails when the sample size is smaller, which happens due to the reduced degrees of freedom (df).

The shape of the Student's t-distribution is determined by the degrees of freedom, which is n - 1 for a sample size n. As n approaches the population size N, the t-distribution approaches the standard normal distribution. This is further supported by the central limit theorem, which states that as the sample size increases, the sampling distribution of the sample means tends to be normal. Therefore:

The statement in the question is false; the t-distribution becomes less peaked and has thicker tails as the degrees of freedom decrease.

The Student's t-distribution is a probability distribution that is used when the sample size is small and the population standard deviation is unknown. It is similar to the standard normal distribution, but has thicker tails and less probability in the center. As the degrees of freedom increase, the t-distribution approaches the standard normal distribution. Therefore, the statement in the question is false. The t-distribution becomes less peaked and has thicker tails as the degrees of freedom decrease.

Just plug in -3 for x and solve

y= -10

Answer:

[tex]\tilde{x}=(-4,-5)[/tex]

Step-by-step explanation:

When we reflect a point [tex]P=(p_x,p_y)[/tex] over the y-axis, it's reflected to the point [tex]\tilde{P}[/tex] with the same y-coordinate as P and whose x-coordinate is minus the x-coordinate of P, hence

[tex]\tilde{P}=(-p_x,p_y).[/tex]

Therefore, in our particular case the point x=(4,-5) is reflected to the point

[tex]\tilde{x}=(-4,-5).[/tex]

f(x+3) = 6x+20

A function is an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).

f(x) = 6x+2

f(x+3) = 6(x+3) + 2

         = 6x + 18 + 2

         = 6x + 20

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Answer:  The required solutions are

[tex](x,y)=(3,2),~(-3,2),~(3,-2),~(-3, -2).[/tex]

Step-by-step explanation:  We are given to solve the following system of equations:

[tex]4x^2+9y^2=72~~~~~~~~~~~~~~~~~~~~(i)\\\\x^2-y^2=5~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

Let us consider that

[tex]x^2=a,~~~~~y^2=b.[/tex]

So, equations (i) and (ii) becomes

[tex]4a+9b=72~~~~~~~~~~~~~~~~(iii)\\\\a-b=5~~~~~~~~~~~~~~~~~~~~(iv)[/tex]

Multiplying equation (iv) by 4 and then subtracting from equation (iii), we get

[tex](4a+9b)-(4a-4b)=72-4\times 5\\\\\Rightarrow 4a+9b-4a+4b=72-20\\\\\Rightarrow 13b=52\\\\\Rightarrow b=4.[/tex]

From equation (iv), we get

[tex]a-b=5\\\\\Rightarrow a-4=5\\\\\Rightarrow a=9.[/tex]

Therefore,

[tex]a=9~~~~~\Rightarrow x^2=9~~~~~\Rightarrow x=\pm3,\\\\b=4~~~~~\Rightarrow y^2=4~~~~~\Rightarrow y=\pm2.[/tex]

Thus, the required solutions are

[tex](x,y)=(3,2),~(-3,2),~(3, -2),~(-3, -2).[/tex]

we have

[tex]P=\frac{KT}{V}[/tex]

Solve for K means clear the variable K

so

Multiply both sides by [tex]V[/tex]

[tex]PV=V\frac{KT}{V}[/tex]

[tex]PV=KT[/tex]

Divide both sides by [tex]T[/tex]

[tex]PV/T=KT/T[/tex]

[tex]K=PV/T[/tex]

therefore

the answer is

[tex]K=PV/T[/tex]

The answer is 15.

15 students earned an 83% or better on the last math test.

60% of 25 students means 60% * 25 = 15 students earned an 83% or better on Ms. Smith's last math test. Therefore, the answer is 15.

Remark

There are two formulas for doing this. I'll use the one that does not involve the slope. I think it will be easier for you to understand.

Step One

Decide which point is (x1,y1)

Since point P is 2/5 the way up from point A, we will call (x1,y1) point A

Step Two

Write the givens.

Point(A) = (-2,-3)

Point(B) = (3,2)

Step Three

Write the formula

x1 + ratio*(x2 - x1), y1 + ratio(y2 - y1)

Sub and solve.

-2 + 2/5 *(3 - - 2) , -3 + 2/5*(2 - -3)

-2 + 2/5(5) , - 3 + 2

-2 + 2 , - 1

P = (0 , - 1) <<<<<< Answer

If this turns out to be incorrect please notify me.

Grace made an error in Step 5 of evaluating the expression 'three less than the quotient of a number squared and nine, increased by eight' when n = 6. The correct evaluation should result in 12, not 7 as she calculated. Hence, the error is in Step 5, where she incorrectly concluded that the expression equals 7.

The question asks us to identify the step at which Grace made an error in evaluating the expression given the description “three less than the quotient of a number squared and nine, increased by eight” when n = 6. Without seeing the exact steps Grace took, we can still solve the expression ourselves.

Firstly, let's translate the description into an algebraic expression using n as our number:

“Three less than the quotient of a number squared and nine” can be expressed as ((n^2)/9) - 3.

Then, we “increase by eight” to get the final expression: ((n^2)/9) - 3 + 8.

If n = 6, we substitute in the value to get: ((6^2)/9) - 3 + 8, which simplifies to (36/9) - 3 + 8 = 4 + 8 = 12.

Thus, the correct answer when evaluating the expression for n = 6 is 12. Grace's error appears to have occurred in Step 5, where she incorrectly evaluated the final sum as 7.

To find the number of miles between the two points, we need to calculate the distance represented by 2 1/2 inches on the map.

Given that the scale states 1 inch represents 60 miles, we can use this information to determine the distance in miles represented by each inch on the map. So, if 1 inch equals 60 miles, then we can write this as the ratio: 1 inch / 60 miles. To find the distance represented by 2 1/2 inches on the map, we can multiply the ratio by the number of inches: [tex](2 1/2 inches) × (1 inch / 60 miles)[/tex]. Simplifying this expression, we get: ([tex]5/2 inches) × (1 inch / 60 miles).[/tex]

Now, we can multiply the numerators and denominators together: [tex](5/2) × (1/60) = 5/120 = 1/24[/tex]. Therefore, [tex]2 1/2 i[/tex]nches on the map represents 1/24 of the actual distance between the two points. Now, to find the actual distance in miles, we can divide 60 miles (representing 1 inch on the map) by 24: 60 miles [tex]/ 24 = 2.5[/tex] miles. Thus, the two points on the map are approximately 2.5 miles apart from each other in actual distance.

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Answer:

[tex](x+3)(x-15)[/tex] is the required factorized form of the given expression.

Step-by-step explanation:

The given equation is:

[tex]x^2-12x-45[/tex], we have to factorized this equation,

Upon solving the above equation, we get

⇒[tex]x^2-15x+3x-45[/tex]

⇒[tex]x(x-15)+3(x-15)[/tex]

⇒[tex](x+3)(x-15)[/tex]

which is the required factorized form of the given expression.