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2 years ago

A tailor used 2 meters of string to make 3 halloween masks. He used _ of a meter for each mask.

Mathematics

1 answer:

swat322 years ago

3 0

One fifth of a meter for each mask
2/10=1/5

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Please help! 10 points

Orlov [11]

<h2>Greetings!</h2>

Answer:

Angle A is 118°

Step-by-step explanation:

All angles in a circle add up to 180, so this means that all the measurements of the three angles add to 180:

x + 2x + 13 + x - 8 = 180

Clean it up:

3x + x + x + 13 -8

5x -5 = 180

To isolate the 5x, you need to move the +5 over to the other side, making it a negative 5:

5x = 180 - 5

5x = 175

Now divide each side by 5 to get the value of 1x:

$\frac{5x}{5} = \frac{175}{5}$

x = 35

So angle A is:

3x + 13

3(35) + 13 = 118

<h3>So angle A is 118°</h3>
<h2>Hope this helps!</h2>

8 0

2 years ago

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Marge cut 16 pieces of tape for mounting pictures on posterboard each piece of tape is 3/8 inches long how much tape did Marge u

daser333 [38]

Answer:

6 inches of tape

Step-by-step explanation:

To find the total amount of tape that Marge used, you can take the measure of each piece ( $\frac{3}{8}$ ) and multiply by the total number of pieces cut (16):

$\frac{3}{8}*\frac{16}{1}=\frac{48}{8}=6inches$

So, Marge used 6 inches of tape altogether.

7 0

2 years ago

Anybody help me please!!!!!!!!

SCORPION-xisa [38]

Answer:

A reflection over the x-axis

Step-by-step explanation:

7 0

2 years ago

A recipe calls for 3/4 cup of flour Sam only has 2/5 of a cup. How much of the recipe can she make?

lilavasa [31]

-0.35 is as much she can make

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3 years ago

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Find the linear approximation of the function g(x) = 3 root 1 + x at a = 0. g(x). Use it to approximate the numbers 3 root 0.95

Virty [35]

Answer:

$L(x)=1+\dfrac{1}{3}x$

$\sqrt[3]{0.95} \approx 0.9833$

$\sqrt[3]{1.1} \approx 1.0333$

Step-by-step explanation:

Given the function: $g(x)=\sqrt[3]{1+x}$

We are to determine the linear approximation of the function g(x) at a = 0.

Linear Approximating Polynomial, $L(x)=f(a)+f'(a)(x-a)$

a=0

$g(0)=\sqrt[3]{1+0}=1$

$g'(x)=\frac{1}{3}(1+x)^{-2/3} \\g'(0)=\frac{1}{3}(1+0)^{-2/3}=\frac{1}{3}$

Therefore:

$L(x)=1+\frac{1}{3}(x-0)\\\\$The linear approximating polynomial of g(x) is:$\\\\L(x)=1+\dfrac{1}{3}x$

(b) $\sqrt[3]{0.95}= \sqrt[3]{1-0.05}$

When x = - 0.05

$L(-0.05)=1+\dfrac{1}{3}(-0.05)=0.9833$

$\sqrt[3]{0.95} \approx 0.9833$

(c)

(b) $\sqrt[3]{1.1}= \sqrt[3]{1+0.1}$

When x = 0.1

$L(1.1)=1+\dfrac{1}{3}(0.1)=1.0333$

$\sqrt[3]{1.1} \approx 1.0333$

7 0

2 years ago