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3 years ago

Expand Your answer should be a polynomial in standard form. (x+3)(x−5)

Mathematics

1 answer:

kvasek [131]3 years ago

5 0

Answer:

x^2 -2x -15

Step-by-step explanation:

xx+x(-5)+3x+3(-5)

=xx-5x+3x-3 times 5(simplify answer)

=x ^2 -2x-15

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Diagram below shows the curve of a quadratic function

uranmaximum [27]

Answer:

q = -8, k = 2.

r = -6.

Step-by-step explanation:

f(x) = (x - p)^2 + q

This is the vertex form of a quadratic where the vertex is at the point (p, q).

Now the x intercepts are at -6 and 2 and the curve is symmetrical about the line x = p.

The value of p is the midpoint of -6 and 2 which is (-6+2) / 2 = -2.

So we have:

f(x) = 1/2(x - -2)^2 + q

f(x) = 1/2(x + 2)^2 + q

Now the graph passes through the point (2, 0) , where it intersects the x axis, therefore, substituting x = 2 and f(x) = 0:

0 = 1/2(2 + 2)^2 + q

0 = 1/2*16 + q

0 = 8 + q

q = -8.

Now convert this to standard form to find k:

f(x) = 1/2(x + 2)^2 - 8

f(x) = 1/2(x^2 + 4x + 4) - 8

f(x) = 1/2x^2 + 2x + 2 - 8

f(x) = 1/2x^2 + 2x - 6

So k = 2.

The r is the y coordinate when x = 0.

so r = 1/2(0+2)^2 - 8

= -6.

3 0

4 years ago

URGENT!!. Find the perimeter and area 9.85 m 15 m 11 m 9m

SIZIF [17.4K]

hope its helps

Step-by-step explanation:

total area = 99m² + 39.4m²

= 138.4m²

6 0

3 years ago

$475 to 25 hours per hour?

azamat

Answer:

$19 per hour

Step-by-step explanation:

475/25 = 19

7 0

3 years ago

If the sum of the legs of a right triangle is 49 inches and the hypotenuse is 41 inches, find the length of the other two sides

german

The length of one leg is x inches.
The sum of the legs is 49 inches, so the length of the other leg is 49-x inches.
The length of the hypotenuse is 41 inches.

Use the Pythagorean theorem:
$(\hbox{one leg})^2 + (\hbox{the other leg})^2=(\hbox{hypotenuse})^2 \\ x^2+(49-x)^2=41^2 \\
x^2+2401-98x+x^2=1681 \\
2x^2-98x+2401=1681 \ \ \ |-1681 \\
2x^2-98x+720=0 \\ \\
a=2 \\ b=-98 \\ c=720 \\ b^2-4ac=(-98)^2-4 \times 2 \times 720=9604-5760=3844 \\ \\
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-98) \pm \sqrt{3844}}{2 \times 2}=\frac{98 \pm 62}{4} \\
x=\frac{98-62}{4} \ \lor \ x=\frac{98+62}{4} \\
x=\frac{36}{4} \ \lor \ x=\frac{160}{4} \\
x=9 \ \lor \ x=40$

$
49-x=49-9 \ \lor \ 49-x=49-40 \\
49-x=40 \ \lor \ 49-x=9$

The lengths of the legs are 9 inches and 40 inches.

4 0

4 years ago

Solve for k : 62=13k-3

cestrela7 [59]