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2 years ago

14

Please help me with this question!!

1 answer:

GarryVolchara [31]2 years ago

6 0

It’s a right angle! which equals to 90° you already have 20° and 8° so you add them together and you get 28. now you have to subtract 90-28 which equals to 62.

ANSWER: 62°

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konstantin123 [22]

${ \qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

$\qquad \sf \dashrightarrow \: \cfrac{9 {}^{(x + 1)} - 3 {}^{2x} }{4 \times 3 {}^{(2x - 1)} }$

$\qquad \sf \dashrightarrow \: \cfrac{3{}^{2(x + 1)} - 3 {}^{2x} }{4 \times 3 {}^{(2x - 1)} }$

$\qquad \sf \dashrightarrow \: \cfrac{3{}^{(2x + 2)} - 3 {}^{2x} }{4 \times 3 {}^{(2x - 1)} }$

here :

${ \sf {3}^{(2x+2)}=({3}^{2x - 1})\sdot (3³)}$

${ \sf {3}^{(2x)}=({3}^{(2x - 1)})\sdot (3¹)}$

$\qquad \sf \dashrightarrow \: \cfrac{3{}^{(2x - 1)}(3 {}^{3} - 3 {}^{1}) }{4 \times 3 {}^{(2x - 1)} }$

[ taking ${ \sf {3}^{(2x - 1)} }$ common here ]

$\qquad \sf \dashrightarrow \: \cfrac{27 {}^{} - 3 {}^{}}{4 }$

$\qquad \sf \dashrightarrow \: \cfrac{24{}^{} {}^{}}{4 }$

$\qquad \sf \dashrightarrow \: 6$

7 0

1 year ago

Read 2 more answers

Solve the inequality.

gulaghasi [49]

$\\ \sf\longmapsto q+12-2(q-22)>0$

$\\ \sf\longmapsto q+12-2q+44>0$

$\\ \sf\longmapsto q-2q+12+44>0$

$\\ \sf\longmapsto -q+56>0$

$\\ \sf\longmapsto -q>-56$

$\\ \sf\longmapsto q$

6 0

3 years ago

The point (-8,-7) is reflected across the x-axis. What quadrant is the new point located in? Group of answer choices Quadrant I

Molodets [167]

Answer:

it is located in quad III

Step-by-step explanation:

8 0

3 years ago

Assuming that a 90° arc has an exact length of , find the length of the radius of the circle.

fiasKO [112]

If the exact length is s, you have
.. s = r*θ . . . . θ = π/2
.. r = s/θ
.. r = 2s/π

Fill in the exact length for "s" in the formula and compute to find the radius.

7 0

3 years ago

What is the distance from L to M?<br> A. 10 units<br> B. 14 units<br> C. 100 units<br> D. 2 units

nignag [31]

Answer: 10 units

Explanation:

5 0

2 years ago