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musickatia [10]
4 years ago
10

-x^3+\frac{1}{5} −x 3 + 5 1 ​

Mathematics
1 answer:
MissTica4 years ago
7 0

Answer:

-2x^3+\frac{256}{5}

Step-by-step explanation:

<em>We</em> need to simplify this expression.

-x^3+\frac{1}{5}-x^3+51

We add similar terms :

-2x^3+\frac{1+5*51}{5}

Simplify :

-2x^3+\frac{256}{5}

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Alyssa was given a gift card for a coffee shop each morning Alyssa uses the gift card to buy one cup of coffee let a represent t
raketka [301]

Answer:20.00

Step-by-step explanation: next time include the table but i managed to find it anyway

5 0
1 year ago
Choose the equation and the slope of the line that passes through (5,-3) and
madam [21]

Answer:

x = 5; the slope is undefined

Step-by-step explanation:

A line perpendicular to the x-axis is a vertical line.

In a vertical line, every point has a different y-coordinate and the same x-coordinate. Since you want a line that is vertical and passes through the point (5, -3), then every point on the line must have x-coordinate 5 no matter what its y-coordinate is. The slope of a vertical line is undefined.

Answer: The equation is x = 5; the slope is undefined

7 0
3 years ago
Solve for x: 3+x=2x=?
Leokris [45]

Answer:

x=3

Step-by-step explanation:

3+x=2x

Subtract x from each side to get the variable on one side

3+x-x=2x-x

3 = x

4 0
4 years ago
Read 2 more answers
"What is the y-intercept of the line that has a slope of- 4 and passes through the point (A)- 10 (B)-6 (C) 8 (D) 10​
jarptica [38.1K]

The y-intercept which is the point at which x = 0 is; A: -10

<h3>How to find the y-intercept?</h3>

The general form of an equation of a line in slope intercept form is;

y = mx + c

where;

m is slope

c is y-intercept

We are given the slope;

m = -4

Now, the line passes through the point 0, -10. Thus;

(y - y1)/(x - x1) = -4

(y + 10)/(x - 0) = -4

y + 10 = -4x

y = -4x - 10

Thus, the y-intercept which is the point at which x = 0 is; y = -10

Read more about y-intercept at; brainly.com/question/1884491

#SPJ1

8 0
2 years ago
Line AB is tangent to circle O at A. The diagram is not drawn to scale.
Arlecino [84]
AO = 21
BC = 14
OC = radius of the circle = AO = 21
∴ OB = OC + CB = 21 + 14 = 35
<span>Line AB is tangent to circle O at A</span>
∴ AB is perpendicular to AO
∴ Δ OAB is a right triangle at A
Applying Pythagorean theorem
∴  OB² = AO² + AB²
∴ AB² = OB² - AO² = 35² - 21² = 1225 - 441 = 784
∴ AB = √784 = 28
<span />
8 0
4 years ago
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