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Mama L [17]
3 years ago
14

Can someone please explain to me how to solve this problem?

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
3 0
Bro that is so easy!!
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Use distributive property to rewrite 9(7 8)
777dan777 [17]
9(7 + 8) = (9 x 7) + (9 x 8) = 63 + 72 = 135
8 0
3 years ago
Here are the first five terms of Fibonacci sequence.
Pepsi [2]

Answer:

a- 32,52

b-20+32=52

Step-by-step explanation:

I think there is only one 4

as following 4,8,12,20

4+8=12

8+12=20

20+12=32

20+32=52

52+32=84

84+52=136

136+84=220

136+220=356

so 4,8,12,20,32,52,84,136,220,356

5 0
3 years ago
Find an expression for a rational function f(x) that satisfies the conditions: a slant asymptote of y = 2x, vertical asymptote a
lesantik [10]

Complete Question

The complete Question is attached below

Answer:

Option D

Step-by-step explanation:

From the question we are told that:

Slant asymptote of y = 2x

Vertical asymptote at x = 1,

Points (0,6)

Generally the Denominator  is give as

With

Vertical Asymptote at

x -1=0

Therefore

Denominator = (x-1)

Generally Slant asympote 2x Gives the Coefficient of the numerator

Therefore

The expression for a rational function f(x) that satisfies the conditions

F(x)=\frac{2x^2-2x-6}{x-1}

Option D

3 0
3 years ago
The synthesis of a large collection of information that contains well-tested and verified hypotheses about certain aspects of th
djyliett [7]

Answer:

it's a B because it's the best answer for this quistion like they are saying IF I do this Then this should happen

6 0
3 years ago
The figure shown below is composed of a semicircle and a non-overlapping equilateral triangle and contains a hole that is also c
Sloan [31]

Check the picture below.

since we know the radius of the larger semicircle is 8, thus its diameter is 16, which is the length of one side of the equilateral triangle.  We also know the smaller semicircle has a radius of 1/3, and thus a diameter of 2/3, namely the lenght of one side of the small equilateral triangle.

now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.

\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}

\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01

3 0
4 years ago
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