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2 years ago

6. Why is the x-coordinate of the vertexof a parabola the mean of the zeros ofthe function?

Mathematics

1 answer:

Leona [35]2 years ago

7 0

Answer:

<h3>The vertex of a parabola is the point where the parabola crosses its axis of symmetry.If the coefficient of the $x^{2}$ term is postive Step-by-step explanation:</h3>

U shape. the coffiecent of the $x^{2}$ term negative.

Parabola: $y=ax^{2} +bx+c\\y=a(x-h)x^{2} +k\\(h,k)$

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-(1+7x)=6(-7-x)+36 <br> Alg 2

olganol [36]

Let's solve your equation step-by-step.

−(1+7x)=6(−7−x)+36

Step 1: Simplify both sides of the equation.

−(1+7x)=6(−7−x)+36

−7x+−1=(6)(−7)+(6)(−x)+36(Distribute)

−7x+−1=−42+−6x+36

−7x−1=(−6x)+(−42+36)(Combine Like Terms)

−7x−1=−6x+−6

−7x−1=−6x−6

Step 2: Add 6x to both sides.

−7x−1+6x=−6x−6+6x

−x−1=−6

Step 3: Add 1 to both sides.

−x−1+1=−6+1

−x=−5

Step 4: Divide both sides by -1.

−x

−1

−5

−1

x=5

Answer:

x=5

7 0

2 years ago

on a scale drawing, 5 metres is represented by 4 centimetres. write the scale as a ratio in it's simplest form.

ryzh [129]

Answer:

<em>The scale factor is 1:125</em>

Step-by-step explanation:

<u>Scaling</u>

Objects can be represented in a reduced or augmented size by using scaling which consists of multiplying or dividing by a constant factor. We use scaling when representing geographic locations on a map.

On a scale drawing, a distance of 5 meters = 500 cm is represented by 4 centimeters. The scale factor is usually represented as a ratio of the smallest to the biggest magnitude:

$\frac{4}{500}=\frac{1}{125}$

The scale factor is 1:125

8 0

3 years ago

An ostrich egg weighs 64 ounces is the weight of the ostrich egg greater than 5 pounds?

IgorLugansk [536]

Answer:

An ostrich egg is 4 pounds, therefore, an ostrich egg is not greater than 5 pounds.

Step-by-step explanation:

Convert the ounces to pound.

1 pound = 16 ounces

Next, divide.

64 ÷ 16 = 4.

4 < 5

So, an ostrich isn't greater than 5 pounds

Have a wonderful day!

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2 years ago

How to integrate ∫㏑x/χ<br> from 1 to e

Natali [406]

$\int_1^e\dfrac{\ln x}{x}dx,~\text{if}~u=\ln x\to x=e^u\to dx=e^udu\\\\\text{So, finding the new limits:}\begin{cases}x=1\to u=\ln1\to u=0\\x=e\to u=\ln e\to u=1\end{cases}\\\\
\int_1^e\dfrac{\ln x}{x}dx=\int_0^1\dfrac{u}{e^u}e^udu=\int_0^1u\,du=\left[\dfrac{u^2}{2}\right]_0^1=\dfrac{1^2}{2}-\dfrac{0^2}{2}=\dfrac{1}{2}\\\\
\boxed{\int_1^e\dfrac{\ln x}{x}dx=\dfrac{1}{2}}$

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2 years ago

Pls help due at 5 slope

Vsevolod [243]

The answer is 4/5 all you do is rise over run

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2 years ago

6. Why is the x-coordinate of the vertexof a parabola the mean of the zeros ofthe function?​

6. Why is the x-coordinate of the vertexof a parabola the mean of the zeros ofthe function?