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kolbaska11 [484]

2 years ago

13

Plz help I need it!!!!!!!!!!!!

1 answer:

AlekseyPX2 years ago

8 0

Answer:

114°

Step-by-step explanation:

$? = \frac{1}{2} (53 \degree + 175 \degree) \\ \\ ? = \frac{1}{2} \times 228 \degree\\ \\ ? = 114 \degree$

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What is 35% of 51 do not round your answer <br> Please help

solniwko [45]

35% of 51 is 17.85. Hope this helped! :)

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3 years ago

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Given f(x) and g(x) = f(x) + k, look at the graph below and determine the value of k.

Anton [14]

Answer:

Given the graph $f(x) = \frac{1}{3}x -2$ and $g(x) = \frac{1}{3}x + 3$

We have to find the value of k;

Since, g(x) = f(x) +k

$\frac{1}{3}x+3 = \frac{1}{3}x-2 +k$

Subtract $\frac{1}{3}x$ from both sides we get;

$3 = -2 +k$

Add 2 to both sides we get;

$3+2 = -2 +k +2$

Simplify:

$5 = k$

or

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Therefore, the value of k = 5

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2 years ago

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Im new to this app can someone help me please and thank u so much

GaryK [48]

Answer:

$3\frac{1}{4}$

Step-by-step explanation:

we have to divide the given distance by 3.

Then we will be able to find the distance between markers.

Given distance is

$9\frac{3}{4} \\=\frac{36+3}{4}\\ =\frac{39}{4}$

Since, there are three markers,

So, we have to divide this distance by 3

So, the distance between markers is-

$\frac{\frac{39}{4}} {3} \\=\frac{39}{4} * \frac{1}{3} \\=\frac{13}{4}$

So, the distance between marker is

$\frac{13}{4}$

Now, we have to convert this into mixed number.

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2 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

<u>case b)</u> $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

<u>case c)</u> $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

<u>case d)</u> $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

$x^{2}-2x-4=0$

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Convert 6 7/20<br> into a decimal

Veseljchak [2.6K]

Answer:

6.35

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