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4 years ago

What value of k makes the statement true? –(k + 12) = 6 + (–12)

Mathematics

1 answer:

Art [367]4 years ago

5 0

The answer is B.if you solve the other sideit will give -6 which is B

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To factor a trinomial you must find the product of "a" and "b" and find the sum of "c"

maksim [4K]

Answer:

true

Step-by-step explanation:

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3 years ago

Order the numbers by value. LEAST to GREATEST 10 √99 √75 9 9.5

beks73 [17]

Answer:

√75, 9, 9.5, √99, 10

Step-by-step explanation:

that's what I got

4 0

3 years ago

5+2x15+1-6/76^3x2+73*0+0.0000273

NemiM [27]

6+2(15)+1-6/(76^2×2)+73×0+0.0000273
6+30+1-6/11552+0+0.0000273
37-0.000519+0.0000273
36.999

Done

6 0

2 years ago

The perimeter of a rectangle is 22 feet, and the area is 24 square feet. Find the length and width.

irina1246 [14]

The length is 8, and the width is 3. Hope it helps! :) If you could vote my answer as the brainliest, that would be awesome! :)

5 0

3 years ago

find the zeros of the quadratic polynomial 4√5x^2-24x-9√5 and verify its relationship between its zeroes and coefficients.......

Bas_tet [7]

Answer:

The zeros of the quadratic polynomial are

$x=\frac{15\sqrt{5}}{10}$ and $x=-\frac{3\sqrt{5}}{10}$

The relationship between its zeroes and coefficients in the procedure

Step-by-step explanation:

step 1

Find the zeros

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$4\sqrt{5}x^{2}-24x-9\sqrt{5}=0$

$a=4\sqrt{5}\\b=-24\\c=-9\sqrt{5}$

substitute in the formula

$x=\frac{-(-24)(+/-)\sqrt{-24^{2}-4(4\sqrt{5})(-9\sqrt{5})}} {2(4\sqrt{5})}$

$x=\frac{24(+/-)\sqrt{-24^{2}-4(4\sqrt{5})(-9\sqrt{5})}} {8\sqrt{5}}$

$x=\frac{24(+/-)\sqrt{1,296}} {8\sqrt{5}}$

$x=\frac{24(+/-)36} {8\sqrt{5}}$

$x=\frac{24(+)36} {8\sqrt{5}}=\frac{15\sqrt{5}}{10}$

$x=\frac{24(-)36} {8\sqrt{5}}=-\frac{3\sqrt{5}}{10}$

step 2

Find the sum of the zeros and the product of the zeros

Sum of the zeros

$(\frac{15\sqrt{5}}{10})+(-\frac{3\sqrt{5}}{10})=\frac{12\sqrt{5}}{10}=\frac{6\sqrt{5}}{5}$

Product of the zeros

$(\frac{15\sqrt{5}}{10})*(-\frac{3\sqrt{5}}{10})=-\frac{9}{4}$

step 3

Verify that

Sum of the zeros= -Coefficient x/Coefficient x²

Coefficient x=-24

Coefficient x²=4√5

substitute

$\frac{6\sqrt{5}}{5}=-(-24)/4\sqrt{5}\\ \\\frac{6\sqrt{5}}{5}=\frac{6\sqrt{5}}{5}$

therefore

the relationship is verified

step 4

Verify that

Product of the zeros= Constant term/Coefficient x²

Constant term=-9√5

Coefficient x²=4√5

substitute

$-\frac{9}{4}=(-9\sqrt{5})/4\sqrt{5}\\ \\-\frac{9}{4}=-\frac{9}{4}$

therefore

the relationship is verified

8 0

3 years ago