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Veseljchak [2.6K]
3 years ago
5

Find the slope between (-4,1) and (-2,-2)

Mathematics
1 answer:
alukav5142 [94]3 years ago
7 0

Answer:

-1.5

Step-by-step explanation:

y=-1.5x-5

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What is the length of leg s of the triangle below? A. 2.2 B. 4 C. 32 D. 4 E. 1.5 F. 16
stiv31 [10]

Answer:

s = 4

Step-by-step explanation:

Note that the triangle has 2 congruent base angles, both 45°

This means the triangle is isosceles with congruent legs, then

s = 4

4 0
3 years ago
A number a decreased by 7 is 2.
umka21 [38]

Answer:

9

Step-by-step explanation:

Let's first convert this to numbers. When a number is decreased by a certain amount, that is the same as saying that something is subtracted from it. Therefore:

a-7=2

Add 7 to both sides:

a-7+7=2+7

a=9

Hope this helps!

7 0
3 years ago
Read 2 more answers
An online video streaming service offers two different plans for unlimited streaming, Plan A has a one-time $25
skelet666 [1.2K]

Answer:

In 6 months (at the fifth month they are equal in price)

Step-by-step explanation:

Set the month as x.

Plan A's equation: 8x + 25

Plan B's equation: 12x + 5

8x + 25 < 12x + 5

Solve.

-4x < -20

x > 5

In 6 months (at the fifth month they are equal in price)

7 0
3 years ago
Chin Woo bought a home for $160,000. He put down 20 percent. The mortgage is at 8 1/2 % for 25 years. By using the table in the
blagie [28]
Down payment
160,000×0.20=32,000

128000=X[((1-(1+0.085)^(-25))/0.085]
Solve for x
X=12507.10
6 0
4 years ago
Read 2 more answers
Let f(x)=x^3-x-1
alexira [117]
f(x) = x^3 - x - 1

To find the gradient of the tangent, we must first differentiate the function.

f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1

The gradient at x = 0 is given by evaluating f'(0).

f'(0) = 3(0)^2 - 1 = -1

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

y = -x + c

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

f(0) = (0)^3 - (0) - 1 = -1

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}
6 0
3 years ago
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