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4 years ago

(4,-2) and (-2,-2) write the equation.

Mathematics

1 answer:

bixtya [17]4 years ago

8 0

the equation is Y=-2/3x+2/3

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Solve for 2: 4x + 188 6x + 216°

Greeley [361]

Steps:

See attachment.

Description:

The first step is to simplify the equation step by step and to simplify an equation you can first multiplying the factors and use the exponent rules to remove the parentheses. After that you need to combine it a terms. Then you will get your answer.

For more steps and graph see the attachment.

Answer: =1890x+216

Hope this helps.

7 0

3 years ago

Read 2 more answers

Mashutka [201]

Answer: 13 minutes

Step-by-step explanation:

( 0 + 2 + 10 + 8 + 20 + 12 + 30 + 7 + 40 + 1)/10 = 130/10 = 13

Hope this helps!

3 0

2 years ago

What is the total resistance of a parallel circuit that has three loads? Load one has a resistance of 6 ohms. Load two has a res

gizmo_the_mogwai [7]

Answer:

The total resistance of these three resistors connected in parallel is $1.7143\Omega$

Step-by-step explanation:

The attached image has the circuit for finding the total resistance. The circuit is composed by a voltage source and three resistors connected in parallel: $R_1=6\Omega$ , $R_2=3\Omega$ and $R_3=12\Omega$ .

First step: to find the source current

The current that the source provides is the sum of the current that each resistor consumes. Keep in mind that the voltage is the same for the three resistors ( $R_1$ , $R_2$ and $R_3$ ).

$I_{R_1}=\frac{V_S}{R_1}$

$I_{R_2}=\frac{V_S}{R_2}$

$I_{R_3}=\frac{V_S}{R_3}$

The total current is:

$I_S=I_{R_1}+I_{R_2}+I_{R_3}=\frac{V_S}{R_1}+\frac{V_S}{R_2}+\frac{V_S}{R_3}=\frac{R_2\cdot R_3 \cdot V_S+R_1\cdot R_3 \cdot V_S+R_1\cdot R_2 \cdot V_S}{R_1\cdot R_2\cdot R_3}$

$I_S=V_S\cdot \frac{R_2\cdot R_3+R_1\cdot R_3+R_1\cdot R_2}{R_1\cdot R_2\cdot R_3}$

The total resistance ( $R_T$ ) is the source voltage divided by the source current:

$R_T=\frac{V_S}{I_S}$

Now, replace $I_S$ by the previous expression and the total resistance would be:

$R_T=\frac{V_S}{V_S\cdot \frac{R_2\cdot R_3+R_1\cdot R_3+R_1\cdot R_2}{R_1\cdot R_2\cdot R_3}}$

Simplify the expression and you must get:

$R_T=\frac{R_1\cdot R_2\cdot R_3}{R_2\cdot R_3+R_1\cdot R_3+R_1\cdot R_2}$

The last step is to replace the values of the resistors:

$R_T=\frac{(6\Omega )\cdot (3\Omega)\cdot (12\Omega)}{(3\Omega)\cdot (12\Omega)+(6\Omega)\cdot (12\Omega)+(6\Omega)\cdot (3\Omega)}=\frac{12}{7}\Omega=1.7143\Omega$

Thus, the total resistance of these three resistors connected in parallel is $1.7143\Omega$

4 0

3 years ago

When a baseball is thrown upward, its height is a function of time. If this function is given by formula h(t)=−16t^2+24t+5, what

nekit [7.7K]

Answer: the maximal height is 14.75 units of distance.

Step-by-step explanation:

We want to find the maximum height of the function

h(t) = -16*t^2 + 24*t + 5

In order to find the maximum, we need to find the value of t where the derivate of h(t) is equal to zero, then we evaluate our original function in that time.

The derivate of h(t) (or the vertical velocity) is:

h'(t) = 2*(-16)*t + 24 = -32*t + 24

we want to find the value of such:

0 = -32*t +24

32*t = 24

t = 24/32 = 0.75

Now we evaluate our height function in that value.

h(0.75) = -16*0.75^2 + 25*0.75 + 5 = 14.75 units

5 0

4 years ago

If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c

horsena [70]

Answer:

Part 1) $cos(A + B) = \frac{140}{221}$

Part 2) $cos(A - B) = \frac{153}{185}$

Part 3) $cos(A - B) = \frac{84}{85}$

Part 4) $cos(A + B) = -\frac{36}{85}$

Part 5) $cos(A - B) = \frac{63}{65}$

Part 6) $cos(A+ B) = -\frac{57}{185}$

Step-by-step explanation:

the complete answer in the attached document

Part 1) we have

$sin(A)=\frac{8}{17}$

$cos(B)=\frac{12}{13}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{8}{17})^2=1$

$cos^2(A)+\frac{64}{289}=1$

$cos^2(A)=1-\frac{64}{289}$

$cos^2(A)=\frac{225}{289}$

$cos(A)=\pm\frac{15}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{15}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{13})^2=1$

$sin^2(B)+\frac{144}{169}=1$

$sin^2(B)=1-\frac{144}{169}$

$sin^2(B)=\frac{25}{169}$

$sin(B)=\pm\frac{25}{169}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{5}{13}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}$

$cos(A + B) = \frac{180}{221}-\frac{40}{221}$

$cos(A + B) = \frac{140}{221}$

Part 2) we have

$sin(A)=\frac{3}{5}$

$cos(B)=\frac{12}{37}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{3}{5})^2=1$

$cos^2(A)+\frac{9}{25}=1$

$cos^2(A)=1-\frac{9}{25}$

$cos^2(A)=\frac{16}{25}$

$cos(A)=\pm\frac{4}{5}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{4}{5}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{37})^2=1$

$sin^2(B)+\frac{144}{1,369}=1$

$sin^2(B)=1-\frac{144}{1,369}$

$sin^2(B)=\frac{1,225}{1,369}$

$sin(B)=\pm\frac{35}{37}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{35}{37}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}$

$cos(A - B) = \frac{48}{185}+\frac{105}{185}$

$cos(A - B) = \frac{153}{185}$

Part 3) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}$

$cos(A - B) = \frac{24}{85}+\frac{60}{85}$

$cos(A - B) = \frac{84}{85}$

Part 4) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}$

$cos(A + B) = \frac{24}{85}-\frac{60}{85}$

$cos(A + B) = -\frac{36}{85}$

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4 0

4 years ago

(4,-2) and (-2,-2) write the equation.​

(4,-2) and (-2,-2) write the equation.