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2 years ago

A sparting goods manufacturer requires

Mathematics

1 answer:

Tpy6a [65]2 years ago

5 0

12 is the correct answer

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Frieda and Joanne are laying a 17 meter track for model trains. They completed one stretch that is 7.44 meters long and another

Dvinal [7]

D is the correct answer. Hopes that helps.

3 0

2 years ago

Would it be<br> A.<br> B.<br> C.<br> D.

gtnhenbr [62]

Answer:

A. is the answer

7 0

1 year ago

Read 2 more answers

Solve for X.<br> X = [?]<br> 5x - 16 X + 10

Aleks [24]

$\boxed{ \sf{ \color{purple}\huge Answer :}}$

$\: \:$

$\large \tt \purple↪ \: \: \: \: \: 5x - 16 = x + 10$

$\: \:$

$\large \tt \purple↪ \: \: \: \: \:5x - x = 10 + 16$

$\:$

$\large \tt \purple↪ \: \: \: \: \: 4x = 26$

$\:$

$\large \tt \purple↪ \: \: \: \: \: x = \cancel\frac{26}{4}$

$\: \:$

$\underline{ \boxed{\large \tt \purple↪ \red{ \: \: \: \: x = \frac{13}{2} }}}{ \large✓}$

$\: \:$

hope helps ~

4 0

1 year ago

Use the diagram showing m || n, as well as the relationships between interior and exterior angles of ΔABC, to answer the questio

nlexa [21]

The measure of angle ∠ACB is 70°. Then the correct option is C.

<h3>What is the triangle?</h3>

A triangle is a three-sided polygon with three angles. The angles of the triangle add up to 180 degrees.

∠ABC + ∠BCA + ∠CAB = 180°

60° + ∠BCA + 50° = 180°

∠BCA = 180° – 110°

∠BCA = 70°

Then the correct option is C.

The complete question is given below.

Use the diagram showing m || n, as well as the relationships between interior and exterior angles of ΔABC, to answer the questions.

The measure of ABC is 60

The measure of BAC is 50

The measure of ACB will be

a. 50

b. 60

c. 70

d. 120

More about the triangle link is given below.

brainly.com/question/25813512

#SPJ1

4 0

1 year ago

Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

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Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$

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Part III

$\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\$

7 0

2 years ago