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3 years ago

Will give Brainliest

Mathematics

1 answer:

ch4aika [34]3 years ago

7 0

Answer:

B. Quadrilateral PQRS is a trapezoid, because there is one pair of equal slopes, and thus one pair of parallel sides.

Step-by-step explanation:

<u>Given:</u>

The slopes of the sides of quadrilateral PQRS are −23, 78, −23, and 1112.

As we see two of the slopes are same, indicating a pair of parallel sides.

The other two slopes are different, so there is only one pair of parallel sides.

This is a definition of a trapezoid.

Correct choice is B

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Find the unknown side length

Westkost [7]

For any right triangle, we can use the Pythagorean Theorem. The Pythagorean Theorem states that for any right triangle, the legs when squared and added together will be equal to the hypotenuse squared.

In mathematical notation: $a^2 + b^2 = c^2$

Where the variables a and b are the legs and the variable c is the hypotenuse.

Because we know the two side lengths of the triangle, we can solve for the unknown side.

We know the length of one of the legs and the hypotenuse.

Plug in the values.

$10^2 + b^2 = 24^2 \\ \\ 
100 + b^2 = 576 \\ \\ 
b^2 = 476 \\ \\ 
$

$b = \sqrt{476}$

So, the square root of 476 is the unknown length.

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3 years ago

If gasoline costs 2.14 per gallon and you drive 500 miles in a car that gets 30 mpg, what is the cost of gasoline

Vlad [161]

Answer:

35.7

Step-by-step explanation:

First, you would calculate how many gallons of gasoline the car gets. To do this, divide your total mileage of 500 miles by the mpg (30). Now you know how many gallons of gasoline will be used by the car. After calculating, you get 16.67. Multiply this by how much it costs per gallon of gasoline, and you get an answer of $35.7.

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3 years ago

When is a protractor preferred to a ruler when finding a measurement

Keith_Richards [23]

When measuring an angle.

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3 years ago

Find all solutions of the equation in the interval [0, 2pi).

natali 33 [55]

Answer:

Step-by-step explanation:

Begin by squaring both sides to get rid of the radical. Doing that gives you:

$sin^2x=1-cosx$

Now use the Pythagorean identity that says

$sin^2x =1-cos^2x$ and make the replacement:

$1-cos^2x=1-cosx$ . Now move everything over to one side of the equals sign and set it equal to 0 so you can factor:

$1-cos^2x+cosx-1=0$ and then simplify to

$cosx-cos^2x=0$

Factor out the common cos(x) to get

$cosx(1-cosx)=0$ and there you have your 2 trig equations:

cos(x) = 0 and 1 - cos(x) = 0

The first one is easy enough to solve. Look on the unit circle and see where, one time around, where the cos of an angle is equal to 0. That occurs at

$x=\frac{\pi }{2},\frac{3\pi}{2}$