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2 years ago

I need help with the unanswered questions asap!!

Mathematics

2 answers:

devlian [24]2 years ago

8 0

Answer:

11. screenshot attached

12.screenshot attached

13. C

14. B

15. D

16. A

Step-by-step explanation:

Mekhanik [1.2K]2 years ago

6 0

Answer:

I’m not sure about 11 or 12 but here is the rest

13. C, 14. B, 15. D, 16. A

Step-by-step explanation:

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The perimeter of a rectangle is 68 ft.

inessss [21]

Let the length be 9x and the width = 8x
Perimeter = 2(l + w) = 2(9x + 8x) = 2(17x) = 34x = 68
x = 68/34 = 2
9x = 9(2) = 18 ft
8x = 8(2) = 16 ft
Therefore, dimensions are 18 ft by 16ft

5 0

3 years ago

A cold front changes the temperature by -3 F each day. If the temperature started at 0 F, what will the temperature be after 5 d

asambeis [7]

Answer:

The temperature falls 3 degrees avery day so you expect to drop 5 times after 5 days. So the temperature will fall by 5*3 = 15 dgrees. If we started at 0 degrees. We will have -15 degrees.

6 0

3 years ago

3(x+4)=5(x-2) I really need help i have no idea how to do it

Oksana_A [137]

answer: $x = 11$

work:

$3(x+4)=5(x-2)$ | distribute 3 and 5 into the parentheses.

$3x + 12 = 5x - 10$ | subtract 5x and move it over to 3x, and solve.

$-2x + 12 = -10$ | subtract 12 and move it over to -10, and solve.

$-2x = -22$ | divide by -2.

$x = 11$ | final answer.

hope this helps! ❤ from peachimin

6 0

3 years ago

Read 2 more answers

Rationalize the denominator of $\frac{\sqrt{32}}{\sqrt{16}-\sqrt{2}}$. The answer can be written as $\frac{A\sqrt{B}+C}{D}$, whe

musickatia [10]

Rationalizing the denominator involves exploiting the well-known difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

We have

$(\sqrt{16}-\sqrt2)(\sqrt{16}+\sqrt2)=(\sqrt{16})^2-(\sqrt2)^2=16-2=14$

so that

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{\sqrt{32}(\sqrt{16}+\sqrt2)}{14}$

Rewrite 16 and 32 as powers of 2, then simplify:

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{\sqrt{2^5}(\sqrt{2^4}+\sqrt2)}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{2^2\sqrt2(2^2+\sqrt2)}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{4\sqrt2(4+\sqrt2)}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{16\sqrt2+4(\sqrt2)^2}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{16\sqrt2+8}{14}$

$\dfrac{\sqrt{32}}{\sqrt{16}-\sqrt2}=\dfrac{8\sqrt2+4}7$

So we have A = 8, B = 2, C = 4, and D = 7, and thus A + B + C + D = 21.

3 0

3 years ago

A rectangular piece of metal is 20 in longer than it is wide. Squares with sides 4 in long are cut from the four corners and the

Misha Larkins [42]

Answer:

Step-by-step explanation:

This is good! The length of the metal is 20 inches longer than the width. But we fold up 4-inch flaps on each side, so the height of the box is 4 inches, and the length and width will each decrease by 8 inches (4 inches on each side). Since the length and width decrease by the same amount, the length will still be 20 inches longer than the width. So the equations would be:

l = w + 20

vol = l x w x h = 1716 in3

1716 = (w + 20)(w)(4)

1716 = 4w2 + 80w

4w2 + 80w - 1716 = 0

4(w2 + 20w - 429) = 0

4(w + 33)(w - 13) = 0

w + 33 = 0, so w = -33

w - 13 = 0, so w = 13

Since the width can't be negative, the width of the box is 13 and the length is 33 (13 + 20).

But remember, the length and width of the piece of metal are each 8 inches longer than the box, so it was 41 inches by 21 inches.

7 0

3 years ago