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Rasek [7]
3 years ago
9

Sylvia decided to purchase 4 points in order to lower her interest rate on her

Mathematics
2 answers:
antoniya [11.8K]3 years ago
6 0

Answer: $ 5 , 4 0 0......

swat323 years ago
4 0
I’m Pretty sure it’s d
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Factor by grouping 4ax-10bx+6ay-15by
Digiron [165]
Check if there is a giggle throughout the full equation, since there isn’t, we group the equation and we get

(4ax-10bx)+(6ay-15by)

Factor out the gcf of the two groups
2x(2a-5b)+3y(2a-5b)

Not sure how to explain this step in words but so do this

(2x+3y)(2a-5b)

The solution is (2x+3y)(2a-5b)
3 0
3 years ago
In a class of 40 students, 15 offer physics, 20 offer history and 3 offer both physics and history. How many student offer neith
Ainat [17]

Answer:

2 students.

Step-by-step explanation:

add together all of the students and then subtract that from 40.

20+15+3 = 38

40-38 = 2

5 0
3 years ago
Read 2 more answers
And triangular pyramid and a triangular prism have congruent bases and the same height the triangular pyramid has a volume of 90
Genrish500 [490]
The volume of prism can find as the volume of pyramid multiplying with 3.

Answer: 90x3 = 270 m³ (volume of prism)
4 0
3 years ago
PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP!!!!
KatRina [158]

Answer:

Give tjbears branliest

Step-by-step explanation:

4 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
3 years ago
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