What is the value of x? show your work

The contrapositive of the proposition 'If the lights are off, then there is no one inside' is 'If there is someone inside, then the lights are not off'. Contrapositives are formed by negating and interchanging the hypothesis and conclusion of a conditional statement, and is a key concept in mathematics.

In logic, the contrapositive of a conditional statement is formed by negating both the hypothesis and the conclusion, and then interchanging the resulting negations. So, the contrapositive of the given proposition 'If the lights are off, then there is no one inside' would be 'If there is someone inside, then the lights are not off'.

This implies that, in the context of this statement, if we know that there is someone inside, we can deduce that the lights must not be off as per the contrapositive. Thus, contrapositives are a helpful way to understand conditional statements from another perspective, and are an integral concept in logic and mathematics.

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Answer: true.

Step-by-step explanation: do you really want an explanation?

joe can only use 3gb of data . bill is 21

21+9+30

1gb=3

3*9=9

3(3)+21=30

The highest common factor of 38 and 34 is 2. Dividing both coefficients by 2 gives us 19d and 17. Therefore, the factored expression of 38d + 34 is 2(19d + 17).

Factoring out the coefficient of a variable in the expression 38d + 34 involves finding a number that divides both coefficients exactly. The highest common factor (HCF) of 38 and 34 is 2. So, we divide 38 and 34 by 2 to get 19d and 17 respectively.

Therefore, the expression, when factored out, becomes 2(19d + 17). This means that the expression 38d + 34 can thus be factored into 2 times the expression (19d + 17).

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Answer:

D. There's a problem with changing the mixed number to an improper fraction, as well as changing the division problem to multiplication.

Step-by-step explanation:

We have shown work of a student to solve a division of fractions. We are asked to check student's work.

Let us check

[tex]2\frac{6}{7}\div\frac{2}{7}[/tex]

Convert mixed fraction into improper fraction:

[tex]2\frac{6}{7}\Rightarrow \frac{2\times 7+6}{7}=\frac{14+6}{7}=\frac{20}{7}[/tex]

[tex]\frac{20}{7}\div\frac{2}{7}[/tex]

Convert division problem to multiplication problem by flipping the 2nd fraction:

[tex]\frac{20}{7}\times\frac{7}{2}[/tex]

[tex]\frac{20}{1}\times\frac{1}{2}[/tex]

[tex]\frac{20}{2}[/tex]

[tex]10[/tex]

Therefore, the student made two mistakes and option D is the correct choice.

the answer is 2/5, a, b

The smallest number of combos that Pizza World can prepare with one slice of cheese and one slice of pepperoni pizza with no slices left over is 80.

The student is asking how many combos of one slice of cheese and one slice of pepperoni pizza Pizza World can prepare with no slices left over given that a cheese pizza has 10 slices and a pepperoni pizza 16 slices. This question is looking for the least common multiple (LCM) of two numbers, which is the smallest number that is a multiple of both numbers. To find the LCM of 10 and 16, we list the multiples of each number until we find a common multiple. The multiples of 10 are 10, 20, 30, 40, 50, 60, ... and the multiples of 16 are 16, 32, 48, 64, 80, .... The smallest common multiple of 10 and 16 is 80. Therefore, we can prepare 80 combos using 8 cheese pizzas (because 8 x 10 = 80) and 5 pepperoni pizzas (because 5 x 16 = 80).

Answer: [tex]y=6-0.5x[/tex]

Step-by-step explanation:

Given  : The initial height of candle = 6 inches

Amount of candle burn each hour = one half an inch per hour.

[tex]=\dfrac{1}{2}\text{ inch per hour}[/tex]

Now , the height of candle = ( initial height ) -( Rate of burning) (Numb er of hours)

Let y be the height of the candle after x hours.

Then, the equation to represent the situation will be :

[tex]y=6-\dfrac{1}{2}x\\\\\Rightarrow\ y=6-0.5x[/tex]

The equation to represent the situation will be, [tex]y=6-\frac{1}{2} \times x[/tex].

Given A candle is 6 inches tall.

The candle burns at a rate of one half an inch per hour.

Let [tex]x[/tex] be the number of hours candle burn and [tex]y[/tex] be the height of the candle after [tex]x[/tex] hours.

Now , the height of candle = ( initial height ) -( Rate of burning) (Number of hours)

So, the equation to represent the situation will be, [tex]y=6-\frac{1}{2} \times x[/tex].

For more details on equation forming follow the link:

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Answer:

D

Step-by-step explanation:

Just did it

the true statements by CPCTC are options A, B, and C. [tex]\( \overline{KL} \cong \overline{ST} \)[/tex] ,[tex]\( \overline{JK} \cong \overline{RS} \)[/tex], [tex]\( \angle K \cong \angle S \)[/tex].

CPCTC stands for "Corresponding Parts of Congruent Triangles are Congruent." It means that if two triangles are congruent, then their corresponding parts are also congruent.

Given that △ JKL ≌ △ RST, let's examine each option:

A. [tex]\( \overline{KL} \cong \overline{ST} \)[/tex] - True by CPCTC, corresponding sides of congruent triangles are congruent.

B. [tex]\( \overline{JK} \cong \overline{RS} \)[/tex] - True by CPCTC, corresponding sides of congruent triangles are congruent.

C. [tex]\( \angle K \cong \angle S \)[/tex] - True by CPCTC, corresponding angles of congruent triangles are congruent.

D. [tex]\( \angle L \cong \angle R \)[/tex] - This statement cannot be determined solely from the fact that △ JKL ≌ △ RST. The corresponding angles may or may not be congruent, depending on the specific congruent triangles.

E. [tex]\( \overline{JL} \cong \overline{RS} \)[/tex] - This statement cannot be determined solely from the fact that △ JKL ≌ △ RST. The corresponding sides may or may not be congruent, depending on the specific congruent triangles.

F. [tex]\( \angle J \cong \angle S \)[/tex] - This statement cannot be determined solely from the fact that △ JKL ≌ △ RST. The corresponding angles may or may not be congruent, depending on the specific congruent triangles.

Therefore, the true statements by CPCTC are options A, B, and C.

The complete question is:

If △ JKL≌ △ RST, which congruences are true by CPCTC? Check all that apply.

A. [tex]\( \overline{KL} \cong \overline{ST} \)[/tex]

B. [tex]\( \overline{JK} \cong \overline{RS} \)[/tex]

C. ∠ K≌ ∠ S

D. ∠ L≌ ∠ R

E. [tex]\( \overline{JL} \cong \overline{RS} \)[/tex]

F. ∠ J≌ ∠ S

Let width be x then

Length is x+8

Therefore area which is given by L*W will be

Area=x*(x=8)

But area =825 sq inches

x*(x+8)=825

x^2+8x-825=0

we use quadratic formula to solve this

and get our x=25 therefore

our width is 25

and length is 33