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3 years ago

Which inequality is true for 50pts

Mathematics

2 answers:

shepuryov [24]3 years ago

8 0

Answer:

B. $(5+2)^{2} > 5^2 + 2^2$

Step-by-step explanation:

The inequality "B" is correct because when you simplify it, you can see that it holds true:

$7^{2} > 25 + 4$

49 > 29

49 is indeed greater than 29, making this inequality true.

enot [183]3 years ago

8 0

Answer:

B. (5+2)^{2} > 5^2 + 2^2

The inequality "B" is correct because when you simplify it, you can see that it holds true:

Step-by-step explanation:

7^{2} > 25 + 4

49 > 29

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Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

3 years ago

Evaluate 33b-40 when b=11.

tester [92]

Answer:

323

Step-by-step explanation:

$33b-40\\\\b=11\\\\33(11)-40\\\\363-40\\\\323$

3 0

3 years ago

1. Describe how the figures are alike. 2. Describe how the figures are different.

IgorLugansk [536]

Some figures have more faces than others.
some have diffrent base

They are all 3-Dimensional shapes

5 0

3 years ago

What’s the slope of the graph? (photo included) giving out brainliest!

ludmilkaskok [199]

Answer:

Using the slope formula with points (0,-1) and (-4,4)

Step-by-step explanation:

Step 1: find the points, (0,-1) and (-4,4)

Step 2: Insert plot points into Slope formula: (tip: M means slope)

m(slope)= $\frac{y_{2} - y_{1}}{x_{2}-x_{1}}$ -> m(slope)= $\frac{4-(-1)}{-4-0}$

Step 3: Solve:

m(slope)= $\frac{5}{-4}$

Answer:

Slope is $\frac{5}{-4}$

or m = $\frac{5}{-4}$

OR (if your teacher wants the answer in decimal form rounded to the nearest 10th.) m = -1.3

6 0

3 years ago

Which point is an x-intercept of the graph of y = 2x² - 7x + 6

AVprozaik [17]

Your answer would be (2,0)

5 0

3 years ago

Which inequality is true for 50pts​

Which inequality is true for 50pts