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3 years ago

Which statement must always be true?

Mathematics

1 answer:

Gennadij [26K]3 years ago

6 0

the statement c will always be true in the end

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suppose a parabola has an axis of symmetry at x = -8, a maximum height of 2, and passes through the point (-7, -1). Write the eq

lesya [120]

Answer:

Step-by-step explanation:

f(x) = a(x - h)² + k - the vertex form of the quadratic function with vertex (h, k)

the axis of symmetry at x = -8 means h = -8

the maximum height of 2 means k = 2

So:

f(x) = a(x - (-8))² + 2

f(x) = a(x + 8)² + 2 - the vertex form of the quadratic function with vertex (-8, 2)

The parabola passing through the point (-7, -1) means that if x = -7 then f(x) = -1

so:

-1 = a(-7 + 8)² + 2

-1 -2 = a(1)² + 2 -2

-3 = a

Threfore:

The vertex form of the parabola which has an axis of symmetry at x = -8, a maximum height of 2, and passes through the point (-7, -1) is:

f(x) = -3(x + 8)² + 2

8 0

3 years ago

What is the product ? (4y-3)(2y^2+3y-5)

yuradex [85]

Answer:

8y^3 + 6y^2 - 29y + 15

Step-by-step explanation:

8y^3 + 12y^2 - 20y - 6y^2 - 9y + 15

then

8y^3 + 6y^2 - 29y + 15 is the product

Best regards

4 0

3 years ago

Read 2 more answers

The Martin family's truck gets an average of 25 miles per gallon. Predict how many miles they can drive using 7 gallons of gas.

victus00 [196]

I believe they can drive 3.57 miles.

8 0

3 years ago

Read 2 more answers

What are the solutions to the following system? StartLayout Enlarged left-brace 1st row negative 2 x squared + y = negative 5 2n

Yakvenalex [24]

Answer:

$(\sqrt{2},-1),(-\sqrt{2},-1)$

(StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1)

Step-by-step explanation:

we have

$-2x^{2} +y=-5$ ----> equation A

$y=-3x^{2} +5$ -----> equation B

solve by substitution

substitute equation B in equation A

$-2x^{2} +(-3x^{2} +5)=-5$

solve for x

$-5x^{2} +5=-5$

$-5x^{2}=-10$

$x^{2}=2$

$x=\pm\sqrt{2}$

Find the value of y

$y=-3x^{2} +5$

For $x=\sqrt{2}$ ----> $y=-3(\sqrt{2})^{2} +5=-1$

For $x=-\sqrt{2}$ ----> $y=-3(-\sqrt{2})^{2} +5=-1$

therefore

The solutions are

$(\sqrt{2},-1),(-\sqrt{2},-1)$

(StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1)

8 0

3 years ago

Read 2 more answers

If, P(x,y) is a point on the unit circle determined by real number δ, then tanδ = what

Lisa [10]

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
=\cfrac{y}{r}
=\cfrac{1}{csc(\theta)}
\\ \quad \\\\
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}
=\cfrac{x}{r}
=\cfrac{1}{sec(\theta)}
\\ \quad \\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
=\cfrac{y}{x}
=\cfrac{sin(\theta)}{cos(\theta)}$

$\bf cot(\theta)=\cfrac{adjacent}{opposite}
=\cfrac{x}{y}
=\cfrac{cos(\theta)}{sin(\theta)}
\\ \quad \\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
=\cfrac{r}{y}
=\cfrac{1}{sin(\theta)}
\\ \quad \\\\
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}
=\cfrac{r}{x}
=\cfrac{1}{cos(\theta)}\\\\
-------------------------------\\\\
P(x,y)\implies \delta\qquad tan(\delta)=\cfrac{y}{x}$

5 0

3 years ago