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2 years ago

Help please not much time!!!!

Mathematics

1 answer:

Rzqust [24]2 years ago

7 0

Answer:

I'm here! This Problem is easy :) The answer is 4

Step-by-step explanation:

2x+10=5x−2

Subtract 5x from both sides

2x+10−5x=−2

Combine 2x and −5x to get −3x.

−3x+10=−2

Subtract 10 from both sides.

−3x=−12

Divide both sides by −3.

Divide −12 by −3 to get 4.

$x=\frac{-12}{-4}$

x = 4 (D)

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PLEASE HELP!!! Will give brainiest!

pentagon [3]

Answer:

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3 years ago

Y+6=2 Make sure all work is shown

defon

Answer:

y=-4

Step-by-step explanation:

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3 years ago

Read 2 more answers

The hypotenuse of a 45°-45°-90° triangle measures 128 cm. A right triangle is shown. The length of the hypotenuse is 128 centime

Aleks [24]

Answer:

64√2 or 64 StartRoot 2 EndRoot

Step-by-step explanation:

A 45-45-90 traingle is a special traingle. Let's say one of the leg of the triangle is x. The other one is also x because of the isosocles triangle theorem. Therefore, using the pytagorean theorem, you find that x^2+x^2=c^2. 2(x)^2=c^2. You then square root both sides and get c= x√2.

Therefore, the two legs are x and the hypotenuse is x√2. x√2=128 because the question says that the hypotenuse is 128. Solve for x by dividing both sides by √2. X=128/√2. You rationalize it by multiplying the numberator and denominator of the fraction by √2. √2*√2= 2.

X=(128√2)/2= 64√2 cm.

Since X is the leg, the answer would be 64√2

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2 years ago

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Two cars start moving from the same point. one travels south at 48 mi/h and the other travels west at 20 mi/h. at what rate is t

timofeeve [1]

Two cars start moving from the same point. one travels south at 48 mi/h and the other travels west at 20 mi/h. at what rate is the distance between the cars increasing four hours later? so its 400%

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2 years ago

Evaluate the definite integral using the graph of f(x) (Image included)

Tanya [424]

a) The first integral corresponds to the area under y = f(x) on the interval [0, 3], which is a right triangle with base 3 and height 5, hence the integral is

$\displaystyle \int_0^3 f(x) \, dx = \frac12 \times 3 \times 5 = \boxed{\frac{15}2}$

b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so

$\displaystyle \int_3^5 f(x) \, dx = \int_3^4 f(x) \, dx + \int_4^5 f(x) \, dx = \int_3^4 f(x) \, dx - \int_3^4 f(x) \, dx = \boxed{0}$

c) The integral over [5, 9] is the negative of the area of a rectangle with length 9 - 5 = 4 and height 5, so

$\displaystyle \int_5^9 f(x) \, dx = -4\times5 = -20$

Then by linearity, we have

$\displaystyle \int_0^9 f(x) \, dx = \left\{\int_0^3 + \int_3^5 + \int_5^9\right\} f(x) \, dx = \frac{15}2 + 0 - 20 = \boxed{-\frac{25}2}$

8 0

2 years ago