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3 years ago

3h-5h+11=17 Solve Equation.

Mathematics

1 answer:

chubhunter [2.5K]3 years ago

3 0

The answer is -3
hope this help!!!!!

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how do you write y - 2 = -(x - 3) in slope-intercept form? A) y = x + 3 B) y = x - 1 C) y = -x + 5 D) y = -3x - 1

SashulF [63]

C) y= -x+5
explain:
y-2=-x+3
y= -x+3+2
so y= -x+5

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2 years ago

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Neporo4naja [7]

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3 years ago

Jeremey performs the same operations on four values of x. He records each resulting y-value in a table, as shown below. Which eq

Anestetic [448]

Answer:

Y=4x-3

Explanation: If we look 2 = 5 We can do 4(2)= 8 -3 = 5 So that the answer and 4(4)= 16-3 = 13 so these are the Y so this is the answer!

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2 years ago

Help Please?<br> thats my image below

kenny6666 [7]

Answer:

6) 7.85 7) 32.97

Step-by-step explanation:

6) p= 3.14×(5/2)

p= 7.85

7) p= 3.14×10×5

p= 32.97

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2 years ago

Verify that -5, 1/2, and 3/4 are the zeroes of the cubic polynomial 4x^3+20x^2+2x-3. Also verify the relationship between the ze

Finger [1]

Solution:

we have been asked to verify that -5, 1/2, and 3/4 are the zeroes of the cubic polynomial $4x^3+20x^2+2x-3$

To verify that whether the given values are zeros or not we will substitute the values in the given Polynomial, if it will returns zero, it mean that value is Zero of the polynomial. But if it return any thing other than zeros it mean that value is not the zero of the polynomial.

Let $f(x)=4x^3+20x^2+2x-3\\\\\text{when x=-5}\\\\f(-5)=4(-5)^3+20(-5)^2+2(-5)-3=-13\\\\\text{when x=}\frac{1}{2}\\$

$f( \frac{1}{2} ) = 4 ( \frac{1}{2} )^3+20(\frac{1}{2})^2+2(\frac{1}{2})-3=\frac{7}{2}\\\\$

$\text{when x=}\frac{3}{4}\\\\$

$f( \frac{3}{4} ) = 4 ( \frac{3}{4} )^3+20(\frac{3}{4})^2+2(\frac{3}{4})-3=\frac{183}{16}\\$

Hence -5, 1/2, and 3/4 are not the zeroes of the given Polynomial.

Since sum of roots $=\frac{-b}{a}= \frac{-20}{4}=-5\\$

But $-5+\frac{1}{2}+\frac{3}{4}= \frac{-15}{4}\neq-5$

Hence we do not find any relation between the coefficients and zeros.

Anyway if the given values doesn't represents the zeros then those given values will not have any relation with the coefficients of the p[polynomial.

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3 years ago