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3 years ago

I suck at math so please help

Mathematics

1 answer:

defon3 years ago

5 0

Answer: c

Step-by-step explanation:

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3) Josie's collection of 39 coins includes twice as many nickels than quarters. If

Degger [83]

Answer:

13 quarters, 26 nickels.

Step-by-step explanation:

n will be number of nickels, q will be number of quarters, and we have 39 total coins

so n + q = 39 (number of nickels + number of quarters = 39 coins.

The problem also says there are twice as many nickels as quarters. so we can call the number of nickels n or we can call it 2q (2 times the number of quarters)

which means

n + q = 39

2q + q = 39

3q = 39

q = 13

and she has 2 times as many nickels as quarters... 2 x 13 = 26 nickels

LMK if you have questions.

If you want to double check the money,

n(0.05) + q(0.25) = 4.55 is the equation to use.

26(0.05) + 13(0.25) = 1.3 + 3.25 = $4.55

5 0

3 years ago

Line segment XY begins at (-6,4) and ends at (-2,4). The segment is reflected over the x-axis and translated left 3 units to for

Marat540 [252]

Line segment XY begins at (-6,4) and ends at (-2,4); XY is reflected over the x-axis and translated left 3 units to form line segment X'Y' which has a length of 4 units.

<h3>What is a system of equations?</h3>

A system of equations is two or more equations that can be solved to get a unique solution. the power of the equation must be in one degree.

Line segment XY begins at (-6,4) and ends at (-2,4). Hence:

$XY = \sqrt{(4-4)^2 + (-2 +6)^2} \\\\XY = \sqrt{0+16} \\\\XY = \sqrt{16} \\\\XY = 4$

XY is reflected over the x-axis and translated left 3 units to form line segment X'Y'. Hence:

XY = X'Y' = 4 units

Line segment XY begins at (-6,4) and ends at (-2,4); XY is reflected over the x-axis and translated left 3 units to form line segment X'Y' which has a length of 4 units.

Find out more on equation at:

brainly.com/question/2972832

#SPJ1

5 0

2 years ago

Prove that if d|m and d|n, then d|(m+n)

Mila [183]

$d\mid m$ means to say there is some $k_1$ such that $m=dk_1$ . Similarly, $d\mid n$ is equivalent to saying there exists $k_2[/te]x such that [tex]n=dk_2$ .

Now, $m+n=dk_1+dk_2=d(k_1+k_2)=dk_3$ , i.e. there is some $k_3$ such that $m+n$ is a multiple of $d$ , i.e. $d\mid(m+n)$ .

5 0

4 years ago

Help pleaseee

Artemon [7]

Answer:

that's it brainliest me

6 0

3 years ago

Find the value of the trigonometric ratio.simply the fraction if needed.

kirill [66]

Answer:

Cos A = Adjacent / hypochondrium

Cos A= 9/ 41

4 0

3 years ago