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4 years ago

8. Solve the inequality –24 ≤ x – 3 – 8x. A. x ≥ 3 B. x ≥ –9 C. x ≤ 3 D. x ≥ –3

Mathematics

2 answers:

kirill [66]4 years ago

8 0

I agree with the same answer with the first answer is c

disa [49]4 years ago

6 0

-24≤x-3-8x

Combine like terms
-24≤-7x-3

Add 3 to both sides
-21≤-7x

Divide both sides by -7. Flip the inequality sign
3≥x

Final answer: C

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The measurements obtained for the interior dimensions of a rectangular box are 200 cm by 200 cm by 300 cm. If each of the three

vlada-n [284]

Answer:

C. 160,000

Step-by-step explanation:

Given that the measurements obtained for the interior dimensions of a rectangular box are 200 cm by 200 cm by 300 cm.

Also given that each of the three measurements has an error of at most 1 cm

COnsider the worst case where each dimension is increased by 1 cm

Then we measure the dimensions as 201 , 201 and 301

So volume would be measured as

$201*201*301\\= 12160701$ cubic cm

Actual volume of the box = $200*200*300\\=12000000$ cubic cm

Difference maximum possible = 160701 cubic cm

Out of the five options given option C is nearest to this value

So answer is

C. 160,000

5 0

3 years ago

B. 243.875 to the nearest hundredth.

WARRIOR [948]

Answer:

the answer is 244 because .875 will round to the next nearest whole number .

Step-by-step explanation:

5 0

4 years ago

U(8, 9) and V(2, 5) are the endpoints of a line segment. What is the midpoint M of that line segment?

Crazy boy [7]

Answer:

(5, 7 )

Step-by-step explanation:

Given 2 points (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

[ 0.5(x₁ + x₂ ), 0.5(y₁ + y₂ ) ]

here (x₁, y₁ ) = U(8, 9) and (x₂, y₂ ) = V(2, 5), thus

midpoint = [0.5(8 + 2), 0.5(9 + 5) ] = [0.5(10), 0.5(14) ] = (5, 7 )

6 0

4 years ago

Which is what is 31 square root to the nearest tenth

stich3 [128]

Hi there!

$\sqrt{31} = 5.567...$
Therefore the answer is about 5.6

7 0

3 years ago

Integrate dx/3sinx+4cosx

german

$\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}$

A standard approach would be the tangent half-angle substitution:

$t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx$

Then

$\sin x=2\sin\dfrac x2\cos\dfrac x2\implies\sin x=\dfrac{2t}{1+t^2}$

$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2\implies\cos x=\dfrac{1-t^2}{1+t^2}$

from which we get

$\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt$

So the integral becomes

$\displaystyle\int\frac{\frac2{1+t^2}}{\frac{6t}{1+t^2}+\frac{4(1-t^2)}{1+t^2}}\,\mathrm dt=\int\frac{\mathrm dt}{3t+2(1-t^2)}=-\int\frac{\mathrm dt}{2t^2-3t-2}$

Rewrite the denominator as

$2t^2-3t-2=(2t+1)(t-2)$

and expand the integrand into its partial fractions:

$\dfrac1{2t^2-3t-2}=\dfrac15\left(\dfrac1{t-2}-\dfrac2{2t+1}\right)$

We have

$\displaystyle-\frac15\int\frac1{t-2}-\frac2{2t+1}\,\mathrm dt=-\frac15(\ln|t-2|-\ln|2t+1|)+C$

$=\dfrac15\ln\left|\dfrac{2t+1}{t-2}\right|+C$

$=\dfrac15\ln\left|\dfrac{2\tan\frac x2+1}{\tan\frac x2-2}\right|+C$

6 0

3 years ago