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2 years ago

10

Find the additive inverse of each rational number. 1.- 1/2 2. 0.25 3. 9/10 4. -0.4

1 answer:

AURORKA [14]2 years ago

5 0

Answer:

i think its 3 9/10 hope this helps u

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What is the coefficient of q<br> (2/3q-3/4) and -1/6q-r)

kirza4 [7]

The coefficient in (2/3q-3/4) is 2/3 and in (-1/6q-r) is -1/6

4 0

3 years ago

Write an equation in point slope and slope intercept form of a line that passes through the given point and

nekit [7.7K]

Answer:

Point slope is ( Y+4) = 1/2(x+3)

Slope intercept is Y = 1/2(x) -5/2

Step-by-step explanation:

For the point slope form.

Given the point as (-3,-4)

And the gradient m = 1/2

Point slope form is

(Y - y1) = m(x-x1)

So

X1 = -3

Y1 = -4

(Y - y1) = m(x-x1)

(Y - (-4)) = 1/2(x -(-3))

( Y+4) = 1/2(x+3)

For the slopes intercept form

Y = mx + c

We can continue from where the point slope form stopped.

( Y+4) = 1/2(x+3)

2(y+4)= x+3

2y + 8 = x+3

2y = x+3-8

2y = x-5

Y = x/2 - 5/2

Y = 1/2(x) -5/2

Where -5/2 = c

1/2 = m

7 0

3 years ago

Find tue volume of following

kompoz [17]

Whereas the basic formula for the area shape is length x width the basic formula for volume is length x width c height

4 0

2 years ago

Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!

LUCKY_DIMON [66]

Answer:

$=8\sqrt{15}b^{\frac{7}{2}}$

Step-by-step explanation:

$\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}$

$=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}$

$=\sqrt{40}b^2\sqrt{24b^3}$

$\sqrt{24b^3}$

$=\sqrt{24}\sqrt{b^3}$

$=\sqrt{24}b^{\frac{3}{2}}$

$=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$\sqrt{2^3}$

$=2^{3\cdot \frac{1}{2}$

$=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}$

$2^{\frac{3}{2}+\frac{3}{2}}$

$=2^3$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}$

$b^{\frac{3}{2}+2}$

$=b^{\frac{7}{2}}$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}$

$=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}$

$=8\sqrt{15}b^{\frac{7}{2}}$

4 0

3 years ago

Suppose an isosceles triangle has two sides of length 42 and 35, and that the angle between these two sides is 120 degree. what

Citrus2011 [14]

Suppose that a=42, b=35, c is unknown, and C = 120°

$c^2 = a^2 + b^2 -2ab(cosC)$

$c^2 = 42^2 + 35^2 -(2*42*35*cos120)

(cos120=-0.5)

c^2 = 1764 + 1225 -(2940*cos120)

c^2 = 2989 -(2940*-0.5)

c^2 = 2989+1470

c^2=4459

c= \sqrt{c^2} 

c=\sqrt{4459}$

3 0

3 years ago