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2 years ago

5

(ab - 9)(ab + 8) Please answer.

1 answer:

WINSTONCH [101]2 years ago

3 0

$\text{Use FOIL method:}\ (a+b)(c+d)=ac+ad+bc+bd\\\\(ab-9)(ab+8)=(ab)(ab)+(ab)(8)+(-9)(ab)+(-9)(8)\\\\=a^2b^2+8ab-9ab-72\\\\=\boxed{a^2b^2-ab-72}$

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I really need answer or I’mma flunk uh oh

irga5000 [103]

Answer:

0u

7u

6u

2u

Step-by-step explanation:

All of these question are just a matter of adding like terms

For the first one we have

7u+(-u)+(-6u)

which is equal to

7u-u-6u

Which I'm sure you can see is equal to 0 or 0u

10/2u+2u

10/2u=5u

5u+2u=7u

6u+0u

0u=0

6u

6u+(-4u)

6u-4u

2u

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2 years ago

Read 2 more answers

In ΔVWX, the measure of ∠X=90°, XW = 36, WV = 85, and VX = 77. What ratio represents the sine of ∠W?

eimsori [14]

<u>Given</u>:

In ΔVWX, the measure of ∠X=90°, XW = 36, WV = 85, and VX = 77.

We need to determine the ratio that represents the sine of ∠W

<u>Ratio of sin of ∠W:</u>

The ratio of sin of ∠W can be determined using the trigonometric ratios.

The ratio of $sin \ \theta$ is given by

$sin \ \theta=\frac{opp}{hyp}$

From the attached figure, the opposite side of ∠W is XV and the hypotenuse of ∠W is WV.

Hence, substituting in the above ratio, we get;

$sin \ W=\frac{XV}{WV}$

Substituting the values, we get;

$sin \ W=\frac{77}{85}$

Thus, the ratio of sine of ∠W is $sin \ W=\frac{77}{85}$

6 0

3 years ago

suppose you make $12 an hour as a lifeguard at the community pool during the summer let h represent the number of hours you work

skad [1K]

12•h
$12x15$

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2 years ago

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m<1=6x and m<3= 120. find the value of x for p to be parallel to q. the diagram is not to scale

lord [1]

M<1 = M<3
these 2 angles are equal for parallel lines
6x=120
X=120/6=20

5 0

3 years ago

Find an m > 0 such that the the equation x^4−(3m+2)x^2+m^2=0 has four real solutions that form an arithmetic sequence.

Aleonysh [2.5K]

Answer:

The value of m is 6.

Step-by-step explanation:

Here, the given equation,

$x^4-(3m+2)x^2+m^2=0$

$x^4+0x^3-(3m+2)x^2+0x+m^2=0$

Let the roots of the equation are a-3b, a-b, a+b and a + 3b, ( they must be form an AP )

Thus, we can write,

$a-3b+a-b+a+b+a+3b=\frac{\text{coefficient of }x^3}{\text{coefficient of }x^4}$

$=\frac{0}{1}=0$

$\implies a=0----(1)$

$(-3b)(-b)+(-b)(b)+(b)(3b)+(3b)(-3b)+(-b)(3b)+(-3b)(b)=\frac{\text{coefficient of }x^2}{\text{coefficient of }x^4}}$

$=\frac{-3m-2}{1}$

$3b^2-b^2+3b^2-9b^2-3b^2-3b^2=-3m-2$

$-10b^2=-3m-2$

$\implies b^2=\frac{3m+2}{10}-----(2)$

$(-3b)(-b)(b)(3b)=\frac{\text{Constant term}}{\text{coefficient of}x^4}= m^2$

$9b^4=m^2$

$9(\frac{3m+2}{10})^2=m^2$

$9(\frac{9m^2+4+12m}{100})=m^2$

$81m^2+36+108m=100m^2$

$-19m^2+108m+36=0$

$19m^2-108m-36=0$

$19m^2-114m+6m-36=0$

$19m(m-6)+6(m-6)=0$

$(19m+6)(m-6)=0$

$\implies m=-\frac{6}{19}\text{ or }m=6$

But m > 0,

Hence, the value of m is 6.

4 0

3 years ago