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3 years ago

13

Solve the inequality. Use algebra to solve the corresponding equation. x^2 - 11x + 30 ≥ 0

1 answer:

raketka [301]3 years ago

4 0

Factor

$(x - 6)(x - 5) \geqslant 0 \\$

If ab>0 then a>0 And b> 0

Then

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Gfern025 i dont know

Kipish [7]

Answer:

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4 0

3 years ago

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Angle θ is in standard position. If tan ⁡ θ < 0 and cos ⁡ θ = − 4/5 , find sin ⁡ θ .

Pani-rosa [81]

Answer:

The value found is.

sin θ = 0.6

Step-by-step explanation:

We know that the value of cosθ is given -4/5

cosθ = -4/5

First, lets finds the angle θ by taking inverse of cosine of -4/5

θ = cos⁻¹ (-4/5)

θ = 143.13°

Substitute the found θ = 143.13° it in tanθ to see if satisfies the equation tanθ < 0:

tanθ < 0

tan(143.13°) < 0

-0.75 < 0

Hence the equation is satisfied.

Which means that θ is equals to 143.13° and can be used to find sinθ

sinθ = sin 143.13°

sin 143.13° = 0.6

6 0

3 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

Need help quick its due at 9

DanielleElmas [232]

Their equal because they both go the same

7 0

2 years ago

HOW DO I FIND VERTEX ON A GRAPH?<br><br>screenshot attached

Tomtit [17]

Answer:

1,9

Step-by-step explanation:

The vertex is the highest point

lol;)

3 0

3 years ago

Read 2 more answers