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3 years ago

Which statements describe this trapezoid?

Mathematics

2 answers:

Annette [7]3 years ago

4 0

All angle are right angles

8_murik_8 [283]3 years ago

4 0

B, because there are only two sides that will never cross. the others will.

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Have an average weight of 16 ounces and a standard deviation of 0.2 ounces. the weights of the sugar bags are normally distribut

Oliga [24]

Standardised z score = (16.075 - 16) / 0.2 = 0.375

which gives a probability of 0.1462 for selecting a package of weight 16.075

the probability of weight being excess of 16.075 = 0.5 - 0.1462 = 03538

Probability of picking 16 in excess of 16-075 = 0.3538^16 = 6.0273 * 10^-8

6 0

3 years ago

**PLS HELP! IT WOULD HELP A LOT, AND ILL GIVE BRAINLIEST!**

Inessa [10]

Answer:

she can buy 3 with a total of six dollars remaining after she bought the three video games

Step-by-step explanation:

200-134 =66

66 -60=3

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2 years ago

Read 2 more answers

A basketball hoop witha circumference of 56.52 inches i need to find the diameter

MA_775_DIABLO [31]

Answer:

Exactly, It would be this: 8.99543738355 . But rounded is 9 or 8.99

5 0

3 years ago

Prove that lines 3x-4y=12 and 3y=12-4x are perpendicular.

Svetllana [295]

Answer:

see explanation

Step-by-step explanation:

If 2 lines are perpendicular then the product of their slopes equals - 1

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Consider the given equations

3x - 4y = 12 ( subtract 3x from both sides )

- 4y = - 3x + 12 ( divide terms by - 4 )

y = $\frac{3}{4}$ x - 3 ← in slope- intercept form

with slope m = $\frac{3}{4}$

3y = 12 - 4x = - 4x + 12 ( divide terms by 3 )

y = - $\frac{4}{3}$ x + 4 ← in slope- intercept form

with slope m = - $\frac{4}{3}$

Then

$\frac{3}{4}$ × - $\frac{4}{3}$ = - 1

Since the product of their slopes = - 1 then the lines are perpendicular

3 0

3 years ago

Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))

Sonja [21]

Answer:

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Let $a=\sin^{-1}(\frac{2}{3})$ .

With some restriction on $a$ this means:

$\sin(a)=\frac{2}{3}$

We need to find $\tan(a)$ .

$\sin^2(a)+\cos^2(a)=1$ is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

$(\frac{2}{3})^2+\cos^2(a)=1$

$\frac{4}{9}+\cos^2(a)=1$

Subtract 4/9 on both sides:

$\cos^2(a)=\frac{5}{9}$

Take the square root of both sides:

$\cos(a)=\pm \sqrt{\frac{5}{9}}$

$\cos(a)=\pm \frac{\sqrt{5}}{3}$

The cosine value is positive because $a$ is a number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ because that is the restriction on sine inverse.

So we have $\cos(a)=\frac{\sqrt{5}}{3}$ .

This means that $\tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}$ .

Multiplying numerator and denominator by 3 gives us:

$\tan(a)=\frac{2}{\sqrt{5}}$

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

$\tan(a)=\frac{2\sqrt{5}}{5}$

Let's continue on to letting $b=\cos^{-1}(\frac{1}{7})$ .

Let's go ahead and say what the restrictions on $b$ are.

$b$ is a number in between 0 and $\pi$ .

So anyways $b=\cos^{-1}(\frac{1}{7})$ implies $\cos(b)=\frac{1}{7}$ .

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of $b$ .

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

$(\frac{1}{7})^2+\sin^2(b)=1$

$\frac{1}{49}+\sin^2(b)=1$

Subtract 1/49 on both sides:

$\sin^2(b)=\frac{48}{49}$

Take the square root of both sides:

$\sin(b)=\pm \sqrt{\frac{48}{49}$

$\sin(b)=\pm \frac{\sqrt{48}}{7}$

$\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}$

$\sin(b)=\pm \frac{4\sqrt{3}}{7}$

So since $b$ is a number between $0$ and $\pi$ , then sine of this value is positive.

This implies:

$\sin(b)=\frac{4\sqrt{3}}{7}$

So $\tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}$ .

Multiplying both top and bottom by 7 gives:

$\frac{4\sqrt{3}}{1}= 4\sqrt{3}$ .

Let's put everything back into the first mentioned identity.

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

$\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}$

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

$\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}$

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

4 0

3 years ago