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3 years ago

6

Find an equation of the parabola that has the indicated vertex and whose graph passes through the given point

1 answer:

vichka [17]3 years ago

3 0

Equation of a parabola with vertex at (2, -1) is
y = a(x - 2)^2 - 1

Using the given point: -3 = a(4 - 2)^2 - 1
-2 = a(2)^2
4a = -2
a = -1/2

Therefore, required equation is
y = -1/2(x - 2)^2 - 1
y = -1/2(x^2 - 4x + 4) - 1
y = -1/2x^2 + 2x - 2 - 1
y = -1/2x^2 + 2x - 3

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Simplify |3 - 11| - (15+ 3+2)2

Natalka [10]

Answer:

-26

Step-by-step explanation:

so, the absolute value of 3 and 11 together is 14.

14-(15+3+2)2

14-(20)2

14-40

-26

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3 years ago

-7c , -6c , -1c , 4c , 0c , 0c , 3c work out the mean temperature

mixas84 [53]

Step-by-step explanation:

(-7)+(-6)+(-1)=-14

(-14)-3=-11

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luda_lava [24]

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3 years ago

The total source voltage in the circuit is 6-3i V. What is the voltage at the middle source

Bezzdna [24]

Answer:

<em>The voltage at the middle source is</em> $(2-4\mathbf{i})\ V$

Step-by-step explanation:

<u>Voltage Sources in Series</u>

When two or more voltage sources are connected in series, the total voltage is the sum of the individual voltages of each source.

The figure shown has three voltage sources of values:

$2 + 6\mathbf{i}$

$a + b\mathbf{i}$

$2 - 5\mathbf{i}$

The sum of these voltages is:

$V_t=4+a+(6+b-5)\mathbf{i}$

Operating:

$V_t=4+a+(1+b)\mathbf{i}$

We know the total voltage is $6-3\mathbf{i}$ , thus:

$4+a+(1+b)\mathbf{i}=6-3\mathbf{i}$

Equating the real parts and the imaginary parts independently:

4+a=6

1+b=-3

Solving each equation:

a = 2

b = -4

The voltage at the middle source is $(2-4\mathbf{i})\ V$

5 0

3 years ago

Pls Help me. There is 20 points to anyone who answers this first.

Alexeev081 [22]

bro doesnt it say the correct answer right there??

6 0

3 years ago