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3 years ago

If y=-2 when x=4 find x when y=5

1 answer:

5 0

OK so put x and y as a fraction which would be -2/4 and then put that equal to 5/x. then multiply 5 by 4 and divide by -2 to solve for x.

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Chain complete 40% of her homework on Monday and 2/3 on Tuesday if she started her homework on Monday what fractional part of he

kotykmax [81]

Answer:

1/5

Step-by-step explanation:

For Monday

The fraction of homework Chain completed =

40% of her homework

= 40% = 40/100

= 2/5

For Tuesday

The amount of home work left =

= 1 - 2/5

3/5

The fraction of homework Chain completed on Tuesday = 2/3

= 2/3 Of 3/5

= 2/3 × 3/5

= 2/5

Let the total complete homework be represented by 1

Hence:

The Fractional part of her homework left for her to complete after Tuesday =

1 - (2/5 + 2/5)

Lowest common denominator is 5

= 1 - (2 + 2/5)

= 1 - (4 /5)

= 1/5

8 0

2 years ago

Your gross weekly salary is $800.00. Your federal income tax deduction is $53. The Social Security tax is 6.2%. The Medicare tax

lord [1]

Make an equation for this find the variables P and t and 200. Them plug the into an equation like y=mx+b.

6 0

3 years ago

Multiple Choice What is the sum of five-sixths + start fraction 2 over 3 end fraction? A. one and one-third B. One and one half

polet [3.4K]

Answer:

Step-by-step explanation:

$\frac{5}{6} +\frac{2}{3} \\\\=\frac{5}{6}+\frac{2*2}{2*3} \\\\=\frac{5}{6}+\frac{4}{6} \\\\=\frac{5+4}{6} \\\\=\frac{9}{6} \\\\=\frac{3*3}{3*2} \\\\=\frac{3}{2}$

$=\frac{2+1}{2} \\\\=\frac{2}{2}+\frac{1}{2} \\\\=1+\frac{1}{2}$

B.One and one half

3 0

3 years ago

Which of the following functions is graphed below?

erastovalidia [21]

Could you please attach the graph itself because the question is not proper

3 0

3 years ago

BD is the angle bisector of

AlexFokin [52]

Note: Let us consider, we need to find the $m\angle ABC$ and $m\angle DBC$ .

Given:

In the given figure, BD is the angle bisector of ABC.

To find:

The $m\angle ABC$ and $m\angle DBC$ .

Solution:

BD is the angle bisector of ABC. So,

$m\angle ABD=m\angle DBC$

$3x=x+20$

$3x-x=20$

$2x=20$

Divide both sides by 2.

$x=\dfrac{20}{2}$

$x=10$

Now,

$m\angle DBC=(x+20)^\circ$

$m\angle DBC=(10+20)^\circ$

$m\angle DBC=30^\circ$

And,

$m\angle ABC=(3x)^\circ+(x+20)^\circ$

$m\angle ABC=(4x+20)^\circ$

$m\angle ABC=(4(10)+20)^\circ$

$m\angle ABC=(40+20)^\circ$

$m\angle ABC=60^\circ$

Therefore, $m\angle DBC=30^\circ,m\angle ABD=30^\circ$ and $m\angle ABC=60^\circ$ .

8 0

2 years ago