All categories

2 years ago

How to solve y = 8x - 9

Mathematics

2 answers:

SIZIF [17.4K]2 years ago

8 0

Answer:

what are you looking for? the value of y or x? or do you need it graphed?

Step-by-step explanation:

Alex Ar [27]2 years ago

6 0

Answer:

-72

Step-by-step explanation:

The answer is -72 because 8 * -9 = -72, I can't explain it to you any other way ;-;

You might be interested in

Find the measure of angle 3

amid [387]

You have not given any attachment and it is lacking information, please resend....... Thank you

8 0

2 years ago

In circle D with m \angle CDE= 76m∠CDE=76 and CD=13CD=13 units find area of sector CDE. Round to the nearest hundredth.

Ksju [112]

Answer:

112.03sq. units

Step-by-step explanation:

The area of a sector = theta/360 * πr²

The area of a sector = 76/360 * 3.14 * 13²

The area of a sector = 76/360 * 530.66

The area of a sector = 40,330.16/360

The area of a sector = 112.03sq. units

This gives the area of the sector

7 0

2 years ago

How can I rewrite this in exponential form and then evaluate it?

Vedmedyk [2.9K]

QUESTION:- EVALUATE THE EXPRESSION

EXPRESSION:-

${( \sqrt[6]{27 {a}^{3} {b}^{4} } )}^{2}$

<h3>ANSWER-></h3>

${( \sqrt[6]{27 {a}^{3} {b}^{4} } )}^{2} \\ {( {27 {a}^{3} {b}^{4} } )}^{ \frac{2}{6} } \\ {({27 {a}^{3} {b}^{4} } )}^{ \frac{ \cancel{2}}{ \cancel6 {}^{ \: 3} } } \\ {(27)}^{ \frac{1}{3} } {a}^{ \frac{3}{3} } {b}^{ \frac{4}{3} } \\{(3)}^{ \frac{ \cancel3}{ \cancel3} } {a}^{ \frac{ \cancel3}{ \cancel3} } {b}^{\frac{3 + 1}{3} } \\3a {b}^{ \frac{ \cancel3}{ \cancel3} } {b}^{ \frac{1}{3} }$ $\\ ANSWER->3ab \sqrt[3]{b} \: \: \: \: \: or \: \: 3a {b}^{ \frac{4}{3} }$

6 0

2 years ago

The following integral requires a preliminary step such as long division or a change of variables before using the method of par

shtirl [24]

Division yields

$\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}$

Now for partial fractions: you're looking for constants a, b, and c such that

$\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}$

$\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a$

which gives a + b = 2, c = 0, and 2a = -7, so that a = -7/2 and b = 11/2. Then

$\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}$

Now, in the integral we get

$\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx$

The first two terms are trivial to integrate. For the third, substitute y = x ² + 2 and dy = 2x dx to get

$\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}$

7 0

1 year ago

There are 7 bags. 2 bags are big. The rest are small.How many bags are small?

siniylev [52]

7-2=5
Ans: 5 bags are small.

6 0

2 years ago