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3 years ago

Help math 10 points pls due in 5 min

Mathematics

2 answers:

Alexandra [31]3 years ago

5 0

Where is the answer choices?

ki77a [65]3 years ago

3 0

Answer:

I don't understand what to do...

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Draw 6 circles.<br> Make 3 groups.<br> Shade 1 group.

krek1111 [17]

Answer:

ummm I think this is obvious

Step-by-step explanation:

2 circles in each group, shade in one group of 2.

7 0

3 years ago

Read 2 more answers

A members-only speaker series allows people to join for $28 and then pay $4 for every event attended. What is the total cost for

wariber [46]

Given:

Joining fee = $28

Fee of each event = $4

To find:

Total cost for someone to attend 4 events.

Solution:

Let the number of events be x and total fee be y.

Fee for 1 event = $4

Fee for x events = $4x

Joining fee remains constant. So, the total fee is

$y=28+4x$

Substitute x=4 in this equation.

$y=28+4(4)$

$y=28+16$

$y=44$

Therefore, total cost of 4 events is $44.

8 0

3 years ago

Gail has eight united states coins. Their total value is 93 cents. What coins does she have?

Kryger [21]

8 coins = $0.93

3 quarters = $0.75
1 dime = $0.10
1 nickel = $0.05
3 pennies = $0.03

3 + 1 + 1 + 3 = 8 coins

$0.75 + $0.10 + $0.05 + $0.03 = $0.93

5 0

4 years ago

Write an equation of a line in point-slope and slope intercept form that passes through the point (1, -8) and is perpendicular t

prisoha [69]

Answer:

11y-2x= -90

Step-by-step explanation:

for the point (1,-8)

x=1 and y= -8

for the equation,

2x -4y= -11

x= -1/m

m= -1/x

x = -1/ (-11/2 )

m= 2/11

so for the equation,

y-y1= m(x- x1)

y- -8=2/11(x-1)

11(y+8)= 2(x-1)

11y +88= 2x-2

11y-2x= -90

3 0

3 years ago

Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$

5 0

4 years ago