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3 years ago

Please help!! 55 points!!

Mathematics

1 answer:

Sedbober [7]3 years ago

5 0

28/80 and then you simplify it by dividing the numerator and denominator by 4 and you get 7/20.

B. 7/20 is the correct answer

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The first three terms of a geometric sequence are shown below.

san4es73 [151]

This is a geometric sequence of the form:

a(n)=(x+3)(-2x)^(n-1) because each term is -2x times the previous term...so

a(8)=(x+3)(-128x^7)

a(8)=-128x^8-384x^7

4 0

3 years ago

Read 2 more answers

The circle below has center C, and its radius is 4 feet. Find the area of the sector with central angle DCE given

garik1379 [7]

Answer:

$area = \frac{8}{3}pie ft^2$

Step-by-step explanation:

Area of sector is given as θ/360*πr²

Where,

θ = central angel of sector, m < DCE = 60°

r = radius = 4 feet

Area of sector = $\frac{60}{360}*pie*4^2$

$area = \frac{1}{6}*pie*16$

$area = \frac{1*16}{6}*pie$

$area = \frac{16}{6}*pie$

$area = \frac{8}{3}pie ft^2$

Area of the sector = 8/3π ft²

5 0

3 years ago

What does n/5 - 3n/10 =1/5

dsp73

I assume you're looking for n.
so, take 2n/10-3n/10=2/10. Making everything have the same denominator is always easier.

This way, you have -n/10=2/10.

n=-2.

8 0

3 years ago

HEEEEELP!!!!!!!!!!!!!11 I WILL MARK BRAINLIEST kx-3y=4 4x-5y=7 k is constant and x and y are variables, for what value of k will

Studentka2010 [4]

Answer:

Step-by-step explanation:

There is no solution for system of equations: $a_1x+b_1y=c_1\\a_2y+b_2y=c_2$

if: $a_1=a_2\quad and\quad b_1=b_2\quad and\quad \bold{c_1\ne c_2}$

so first, we we need to transform the equations to the form where the coefficients at y will be the same:

$kx-3y=4\\ 4x-5y=7\\\\ (kx-3y)\cdot5=4\cdot5\\ (4x-5y)\cdot3=7\cdot3\\\\ 5kx-15y=20\\ 12x-15y=21$

Now we have b₁=b₂ and c₁≠c₂ so the system has no solution if a₁=a₂

5k = 12

÷5 ÷5

k = ¹²/₅

7 0

3 years ago

Multiply the sum of 7/8 and 3/4 by the differences of 5/36 from 11/13

Anni [7]

Answer:

$=-\frac{331}{288}=-1.14930\dots$

Step-by-step explanation:

$(\frac{7}{8} + \frac{3}{4} )\times (\frac{5}{36} - \frac{11}{13} )\\\\\mathrm{Join}\:\frac{7}{8}+\frac{3}{4}:\quad \frac{13}{8}\\\\=\frac{13}{8}\left(\frac{5}{36}-\frac{11}{13}\right)\\\\\mathrm{Join}\:\frac{5}{36}-\frac{11}{13}:\quad -\frac{331}{468}\\\\=\frac{13}{8}\left(-\frac{331}{468}\right)\\\\=-\frac{13}{8}\times\frac{331}{468}\\\\\mathrm{Cross-cancel\:common\:factor:}\:13\\=\frac{1}{8}\times\frac{331}{36}\\\\=-\frac{1\cdot \:331}{8\times\:36}\\\\=-\frac{331}{288}$

4 0

3 years ago