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djyliett [7]
3 years ago
6

Subtract (x^2-5x+4)-(8x-9)

Mathematics
2 answers:
cluponka [151]3 years ago
6 0

Answer: x^2 - 13x + 13

Step-by-step explanation:


Sladkaya [172]3 years ago
4 0

Answer:

x^2-13x+13

Step-by-step explanation:

(x^2-5x+4)-(8x-9)=(x^2-5x+4)-8x+9

                           =x^2-5x-8x+4+9

                           =x^2-13x+13

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HELPP!!!!!!!!!!!!!!!!
Anvisha [2.4K]

Answer:

Step-by-step explanation:

We'll take this step by step.  The equation is

8-3\sqrt[5]{x^3}=-7

Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time.  First thing is to subtract 8 from both sides:

-3\sqrt[5]{x^3}=-15

The goal is to isolate the term with the x in it, so that means that the -3 has to go.  Divide it away on both sides:

\sqrt[5]{x^3}=5

Let's rewrite that radical into exponential form:

x^{\frac{3}{5}}=5

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

(x^{\frac{3}{5}})^{\frac{5}{3}}=5^{\frac{5}{3}}

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely.  On the right, let's rewrite that back in radical form to solve it easier:

x=\sqrt[3]{5^5}

Let's group that radicad into groups of 3's now to make the simplifying easier:

x=\sqrt[3]{5^3*5^2} because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:

x=5\sqrt[3]{5^2} which is the same as:

x=5\sqrt[3]{25}

8 0
3 years ago
Determine any data values that are missing from the tablo, assuming that the data represent a linear function.
mina [271]

Answer:

B.

Step-by-step explanation:

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3 years ago
Locate 11/3 on a number line
geniusboy [140]
11/3 would be closlely to 3.5 because it is 3.66667
4 0
3 years ago
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Find a vector equation and parametric equations for the line through the point (1,0,6) and perpendicular to the plane x+3y+z=5.
tester [92]

The normal vector to the plane <em>x</em> + 3<em>y</em> + <em>z</em> = 5 is <em>n</em> = (1, 3, 1). The line we want is parallel to this normal vector.

Scale this normal vector by any real number <em>t</em> to get the equation of the line through the point (1, 3, 1) and the origin, then translate it by the vector (1, 0, 6) to get the equation of the line we want:

(1, 0, 6) + (1, 3, 1)<em>t</em> = (1 + <em>t</em>, 3<em>t</em>, 6 + <em>t</em>)

This is the vector equation; getting the parametric form is just a matter of delineating

<em>x</em>(<em>t</em>) = 1 + <em>t</em>

<em>y</em>(<em>t</em>) = 3<em>t</em>

<em>z</em>(<em>t</em>) = 6 + <em>t</em>

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3 years ago
What is f(-6)<br><br> A. -8<br><br> B. -6<br><br> C. 6<br><br> D. 8
Step2247 [10]

f(x) = y.

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Answer: C

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3 years ago
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