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3 years ago

Help me pLease!!!!!!!!!!!!

Mathematics

2 answers:

Troyanec [42]3 years ago

5 0

Add 3 on both sides: 2I5-2xI≤18
divide 2 on both sides: I5-2xI≤9
-9≤5-2x≤9
subtract 5: -14≤-2x≤4
divide by -2. Remember: when divided by a negative number, the inequality symbol changes direction. so we have7 ≥x≥-2
rewrite it: -2≤x≤7
so the answer is C
you were right.

nikitadnepr [17]3 years ago

3 0

The answer is C
Hope I helped

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A CD has a 5% chance of being a smash hit and profiting $5.2 million, a 50% chance of being a modest success and profiting $0.9

Zinaida [17]

The probability of profiting $5.2 million is 5% = 0.05

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The probability of breaking even, meaning no profit gain, is 45% = 0.45

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3 years ago

Convert equation from slope intercept form to standard form y=-7/3x+1

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Answer:

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2 years ago

What is the value of x

pochemuha

This is the isosceles triangle. Therefore, the angles at the base are congruent.

We know that the sum of the measures of the angles in a triangle is 180°.

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$38^o+x^o+x^o=180^o$

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2 years ago

Joe began to increase the speed of his car at 2:00 P.M. Since that time,the speed of Joes car has been steadily increasing by 1

Olegator [25]

Answer:

7 mins

Step-by-step explanation:

Current speed of Joes Car = 65.5 mph

We have to find the time interval for which the car exceeded the speed limit of 55 mph.

While, we are given that the speed of the car was constantly increasing, hence the speed over all increased from the limit of 55 mph = 65.50-55.00 = 10.50 mph

We are also given that Joes car was increasing speed at a constant rate of 1.50 mph for every passing minute. Hence

1.50 mph is increased in 1 minute

1 mph will be increase in $\frac{1}{1.5}$ minutes

Hence

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$= \frac{10.5}{1.5}$

$=7$

Hence joes car was exceeding the limit of 55 mph for 7 minutes.

8 0

2 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

8 0

3 years ago