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3 years ago

Hello pls help meeee

Mathematics

1 answer:

lakkis [162]3 years ago

5 0

Answer:

answer is below

Step-by-step explanation:

0 1

2 3

5 6

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kkurt [141]

35/42 =5/6

so the answer is 5/6.

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2 years ago

The sum of twice a number and 15 is grater the 3

NNADVOKAT [17]

Answer:

x>-6

Step-by-step explanation:

The expression you need to solve for this is: 2x+15>3

For now pretend that the > is a = and solve as you normally would for x.

Subtract 15 from both sides

2x=-12

Divide by 2

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Put the > back in

x>-6

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3 years ago

Composition of functions Find g (f(x)) for the following problems

beks73 [17]

$f(x)=-3x\qquad g(x)=5x-6\\\\g(f(x))=5(-3x)-6=\boxed{-15x-6}$

$f(x)=x-6\qquad g(x)=x^2+6x-2\\\\g(f(x))=(x-6)^2+6(x-6)-2=x^2-12x+36+6x-36-2=\\\\=\boxed{x^2-6x-2}$

$f(x)=4x+3\qquad g(x)=2x^2-x+1\\\\g(f(x))=2(4x+3)^2-(4x+3)+1=2(16x^2+24x+9)-4x-3+1=\\\\=32x^2+48x+18-4x-3+1=\boxed{32x^2+44x+16}$

$f(x)=2x^2\qquad g(x)=8x^2+3x\\\\g(f(x))=8(2x^2)^2+3(2x^2)=8\cdot4x^4+6x^2=\boxed{32x^4+6x^2}$

7 0

3 years ago

What is the exact distance from (-5,1) to (3,0).

Ilya [14]

Answer:

The distance between the two points is $\sqrt{65} \ \text{units}.$

Step-by-step explanation:

In order to find the distance between two coordinate pairs, we can use the distance formula:

$\displaystyle \bullet \ \ \ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Our coordinate pairs need to be labeled accordingly, so we can use this naming system:

$\bullet \ \ \ (x_1, y_1), (x_2, y_2)$

This assigns a name to our points:

$x_1 = -5$
$y_1 = 1$
$x_2 = 3$
$y_2 = 0$

Therefore, we can plug these into the formula and solve:

$d=\sqrt{(3-(-5))^2+(0-1)^2}\\\\d=\sqrt{(8)^2+(-1)^2}\\\\d=\sqrt{8^2+1^2}\\\\d=\sqrt{64+1}\\\\d=\sqrt{65}$

Therefore, the distance between the two points is $\sqrt{65} \ \text{units}$ .

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2 years ago

Ted is comparing the temperatures of three days in January . The temperatures on Monday and Tuesday were opposites . The tempera

SOVA2 [1]

Tuesday, Wednesday, monday

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3 years ago