3 years ago

Solve for d 3(d+12)=8-4d ?????

Mathematics

1 answer:

zysi [14]3 years ago

5 0

Answer:

d=-4

Step-by-step explanation:

the answer is negative 4 have a nice day :)

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-3(x - 2) - 2x + 4x^2

Fantom [35]

Answer:

-6x+6+4x^2

Step-by-step explanation:

Distribute the -3

-3(x-2)= -3x+6

Combine like terms

<u>-3x</u>+6<u>-2x</u>+4x^2

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3 years ago

F(x)= sqrt of 5x+7 and g(x)= the sqrt of 5x-7

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F(x) = √(5x + 7)

g(x) = √(5x - 7)

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(f + g)(x) = √(5x + 7) + √(5x - 7)

Answer: C)

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3 years ago

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Answer:

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3 years ago

Six and four fifths divided by three and two ninths

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Its going to be 4 .............

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3 years ago

Which equation represents a circle that contains the point (-5, 3) and has a center at (-2, 1)? Distance formula: vaa -02 (x - 1

natali 33 [55]

Given:

The center of the circle = (-2,1).

Circle passes through the point (-5,3).

To find:

The equation of the circle.

Solution:

Radius is the distance between the center of the circle and any point on the circle. So, radius of the circle is the distance between the points (-2,1) and (-5,3).

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

$r=\sqrt{(-5-(-2))^2+(3-1)^2}$

$r=\sqrt{(-5+2)^2+(2)^2}$

$r=\sqrt{(-3)^2+(2)^2}$

On further simplification, we get

$r=\sqrt{9+4}$

$r=\sqrt{13}$

The standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where, (h,k) is the center of the circle and r is the radius of the circle.

Substitute h=-2, k=1 and $r=\sqrt{13}$ .

$(x-(-2))^2+(y-1)^2=(\sqrt{13})^2$

$(x+2)^2+(y-1)^2=13$

Therefore, the equation of the circle is $(x+2)^2+(y-1)^2=13$ .

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3 years ago