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Thepotemich [5.8K]
2 years ago
6

WILL GIVE BRANLIEST

Mathematics
1 answer:
arsen [322]2 years ago
6 0

Answer:

<h2>m<ABD = 36°</h2><h2>m<DBC = 54°</h2>

Step-by-step explanation:

2 + 3 = 5

90° : 5 = 18

18 × 2 = 36

m<ABD = 36°

18 × 3 = 54

m<DBC = 54°

<h3>#Let'sLearn</h3>
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....pls help me :((((((
Rina8888 [55]

Answer:

y= -7

Step-by-step explanation:

So first distribute the 8

4=8y+32-4y

now combine like terms (the -4y+8y)

4=32+4y

Subtract 32 on both sides to isolate 4y

-28=4y

now divide 4 on both sides to isolate Y

-7=y

so y= -7

5 0
3 years ago
Form a paragraph proof for the following. Make sure you include all the
timama [110]

Answer:

Line BD is Congruent to Line BD due to the Reflexive property. Angle Bad is congruent to angle BCD because of the third angle theorem therefore triangle ABD is congruent to triangle CBD because all angles and sides are congruent

I don't know if that will satisfy your teacher because some prefer different things but that leaves no ground to be inferred and explains everything word for word.

8 0
3 years ago
PLEASEEEEE HELPPPPPPP
Vesnalui [34]

1) We are given points on the line of the graph (0,3) and (1,1).

Rise/run represents the slope of the line.

Rise/run = -2/1  = -2.

Therefore, slope of the line is -2.

<h3>Correct option is A) -2 .</h3>

2) Given equation y=7.5x-5.

From the graph we can see that line crosses x-axis at 0.667.

But we don't find any such option.

But we can see line crossing y-axis is -5.

It seems that we need to find the point where line cut y-axis.

<h3>Therefore, correct option is C option.</h3><h3>C) - 5</h3>

6 0
3 years ago
A circle has centre (3,0) and radius 5. The line y = 2x + k intersects the circle in two points. Find the set
lara [203]

A circle with center (3,0) and radius 5 has equation

(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x  = 16

If we substitute y=2x+k in this equation, we have

x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

\Delta=b^2-4ac=(4k-6)^2-4\cdot 5\cdot (k^2-16)>0

The expression simplifies to

-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89

The solutions to the associated equation are

k^2 + 12 k - 89=0 \iff k=-6\pm 5\sqrt{5}

So, the parabola is negative between the two solutions:

-6-5\sqrt{5}

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=2x%20%7B%7D%5E%7B2%7D%20%20%2B%204x%20-%2021%20%3D%200" id="TexFormula1" title="2x {}^{2} + 4
Ugo [173]
The answer should be  21/8 hope it helps

7 0
3 years ago
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