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1 year ago

In a certain city, the percent of children under

1 answer:

3 0

As per given data of living children under 18 years of age , percent of children whose age is under 18 years lived with their mother only is equal to 39.3%.

As given in the question,

Let us consider 100% be the total percent representing the living status of children under 18 years of age

Given:

Percent of children living with both the parents = 49.2%

Percent of children living with only father = 8%

Percent of children living with neither of the parents = 3.5%

Let us consider percent of children living with only mother = x%

Required equation is:

( x + 49.2 + 8 + 3.5 )% = 100%

⇒ x + 60.7% = 100%

⇒ x = 100 - 60.7

⇒ x = 39.3%

Therefore, for the given data of living children under 18 years of age , percent of children whose age is under 18 years lived with their mother only is equal to 39.3%.

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3 years ago

100 points! Mhanifa can you please help? Look at the picture attached. I will mark brainliest!

dimulka [17.4K]

Answer:

1 and 2.

Midpoints calculated, plotted and connected to make the triangle DEF, see the attached.

D= (-2, 2), E = (-1, -2), F = (-4, -1)

As per definition, midsegment is parallel to a side.

Parallel lines have same slope.

Find slopes of FD and CB and compare.

m(FD) = (2 - (-1))/(-2 -(-4)) = 3/2
m(CB) = (1 - (-5))/(1 - (-3)) = 6/4 = 3/2
As we see the slopes are same

Find the slopes of FE and AB and compare.

m(FE) = (-2 - (- 1))/(-1 - (-4)) = -1/3
m(AB) = (1 - 3)/(1 - (-5)) = -2/6 = -1/3
Slopes are same

Find the slopes of DE and AC and compare.

m(DE) = (-2 - 2)/(-1 - (-2)) = -4/1 = -4
m(AC) = (-5 - 3)/(-3 - (-5)) = -8/2 = -4
Slopes are same

As per definition, midsegment is half the parallel side.

We'll show that FD = 1/2CB

FD = $\sqrt{(2+1)^2+(-2+4)^2}$ = $\sqrt{3^2+2^2}$ = $\sqrt{13}$
CB = $\sqrt{(1 + 5)^2+(1+3)^2}$ = $\sqrt{6^2+4^2}$ = 2 $\sqrt{13}$
As we see FD = 1/2CB

FE = 1/2AB

FE = $\sqrt{(-4+1)^2+(-1+2)^2}$ = $\sqrt{3^2+1^2}$ = $\sqrt{10}$
AB = $\sqrt{(-5 -1)^2+(3-1)^2}$ = $\sqrt{6^2+2^2}$ = 2 $\sqrt{10}$
As we see FE = 1/2AB

DE = 1/2AC

DE = $\sqrt{(-2+1)^2+(2+2)^2}$ = $\sqrt{1^2+4^2}$ = $\sqrt{17}$
AC = $\sqrt{(-5 +3)^2+(3+5)^2}$ = $\sqrt{2^2+8^2}$ = 2 $\sqrt{17}$
As we see DE = 1/2AC

3 0

3 years ago

Read 2 more answers

Suppose m<3 = 105. find m<6

tatyana61 [14]

If you mean the m^3=105, then m^6 will be $( \sqrt[3]{105}) ^6$ which is 105^2 which is 11025

7 0

3 years ago

Solve for x using desmos or the quadratic formula 2x^2 - 5x - 12 = 0

ivanzaharov [21]

Answer:

x = 4 or x = (-3)/2

Step-by-step explanation:

Solve for x over the real numbers:

2 x^2 - 5 x - 12 = 0

x = (5 ± sqrt((-5)^2 - 4×2 (-12)))/(2×2) = (5 ± sqrt(25 + 96))/4 = (5 ± sqrt(121))/4:

x = (5 + sqrt(121))/4 or x = (5 - sqrt(121))/4

sqrt(121) = sqrt(11^2) = 11:

x = (5 + 11)/4 or x = (5 - 11)/4

(5 + 11)/4 = 16/4 = 4:

x = 4 or x = (5 - 11)/4

(5 - 11)/4 = -6/4 = -3/2:

Answer: x = 4 or x = (-3)/2

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3 years ago

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In a photograph, a house is 4 inches wide and 6 inches tall. The photograph is enlarged while keeping proportional dimensions, a

AlekseyPX

Since you keep proportional dimensions, the proportion between the old and new dimensions must be the same. So, if we call the new height $h$ , the preservation of the width/height ratio is written as

$4 \div 6 = 9 \div h$

Solving the proportion for $h$ yields

$h = \frac{6\cdot 9}{4} =\frac{3\cdot 9}{2} = \frac{27}{2} = 13.5$

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3 years ago

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