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djverab [1.8K]
3 years ago
11

Helppp this is due in an hour!!​

Mathematics
1 answer:
STatiana [176]3 years ago
4 0
First you solve for the area of the unshaded triangle. Use the formula of area of triangle which is 1/2bh to find. 1/2(9)(4) = 18 cm^2. The area of the unshaded triangle is 18cm^2. Now we need to find the area of the rectangle. We can find that by using the formula for area of rectangle which is (b)(h). 12*4 = 48 cm ^2. Now that we have the area of the triangle and the area of the rectangle, we just need to subtract the area of the triangle from the area of the rectangle to get the area of the shaded region. 48 cm^2 - 18 cm^2 = 30 cm^2.

So our answer is: The area of the shaded region is 30 cm^2. In this case it’s option C or the third option for your quiz.
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I don't get how to work out the repeating problems, could someone explain please?
barxatty [35]
Detecting...

Your answer would be:

0.025 = 1/40 

0.025/1 * 1000/1000 = 25/1000

25 \div 25 / 1000 \div 25 = 1/40

1/40 is your answer.
3 0
3 years ago
Drag the simplified value into the box to match each expression.
AveGali [126]

Answer:

2^{3} +1⋅4−3 = 9

2^{0}+10−4⋅2 = 3

3+5^{2} -6⋅4 = 4

Step-by-step explanation:

4 0
3 years ago
2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.
likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4  + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

y = ^+_- \sqrt{-x^2+4x+4} -5

Thus

f(x) = \sqrt{-x^2+4x+4} - 5

g(x) = -\sqrt{-x^2+4x+4} - 5

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0

The quadratic has roots

\frac{52 ^+_-\sqrt{1664}}{26}

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

5 0
3 years ago
This diagram was created by starting with points Aand
Anton [14]
Sorry I just need the points
7 0
3 years ago
Solve for x: |x − 2| + 10 = 12 (1 point)
Mrrafil [7]

Answer::

A. x= 0 and x= 4

8 0
3 years ago
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