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Fittoniya [83]
2 years ago
11

Please help 7th grade math

Mathematics
1 answer:
Leokris [45]2 years ago
3 0

Answer:

d = 67.5

e = 22.5

Step-by-step explanation:

d + e = 90

d = 3e

Therefore, we can simply divide 90 by 4, which gives us 22.5, and also tells us what e equals. Then, multiply that by 3 to get d.

Hope that this helps!

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A basket of fruit contains four oranges five apples and six bananas. If you choose a piece of fruit at random from the basket, w
Leni [432]

Answer:

6/15

Theres 6 bananas & 15 fruits total

8 0
3 years ago
If tan x =3/4, and 0
jasenka [17]

Answer: x=0.6435011

Step-by-step explanation:

Take the inverse tangent of both sides of the equation to extract  

x

from inside the tangent.

x

=

arctan

(

3

4

)

Evaluate  

arctan

(

3

4

)

.

x

=

0.6435011

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from  

π

to find the solution in the fourth quadrant.

x

=

(

3.14159265

)

+

0.6435011

Simplify the expression to find the second solution.

Tap for more steps...

x

=

3.78509376

Find the period.

Tap for more steps...

π

The period of the  

tan

(

x

)

function is  

π

so values will repeat every  

π

radians in both directions.

x

=

0.6435011

+

π

n

,

3.78509376

+

π

n

, for any integer  

n

Consolidate the answers.

x

=

0.6435011

+

π

n

, for any integer  

n

6 0
2 years ago
Hence, or otherwise, simplify fully (x² + 4) - (x - 2)2
raketka [301]

Answer:

fully simplified answer: x^2-2x+8

Step-by-step explanation:

6 0
3 years ago
The coordinates of the endpoints of AB and CD are A(2,
Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

m= \frac{y_2-y_1}{x_2-x_1}

Formula to Find Distance between two points

D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

mAB ( represents , Slope of AB )

1. mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}

2. mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}

3. mCD=\frac{5-2}{6-5}=\frac{3}{1}=3

4. AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}

5. BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}

mAC = mBC  , and C is common point , hence these three are collinear points  making a straight line whole slope is -\frac{1}{3}

mAB=-\frac{1}{3}

mCD=3

mAB \times mCD = -\frac{1}{3} \times 3 = -1

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0
3 years ago
Given ∆ABC≅∆PQR, mb=8v-9 , find m 31°<br> b.<br> 52°<br> c.<br> 49°<br> d.<br> 32°
Ipatiy [6.2K]
Here, 5v+6 = 8v-9
8v - 5v = 6 + 9
3v = 15
v = 15/3
v = 5

So, m<b = 5(5) + 6 = 25 + 6 = 31
As they are congruent, m<q would be same [ 31 degree ]

In short, Your Answer would be Option A

Hope this helps!
3 0
2 years ago
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