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konstantin123 [22]
3 years ago
15

Draw a model of the expression 3x + 5 please help ;_;

Mathematics
1 answer:
artcher [175]3 years ago
3 0
For 3x + 5:
5 is the y-intercept and 3(also 3/1) is the rise over run

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I’m pretty sure that says 6 cups of vegetables so 84 cups but if you meant 1/6 of a cup then 7/3 cups or 2/1/3
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Any Number that is divisible by four is also divisible by eight
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Answer:

The answer is 16 .

Step-by-step explanation:

Hope you got it

4 0
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

5 0
2 years ago
Read 2 more answers
How do I solve this problem?
Anvisha [2.4K]

Answer:

y+11=-\frac{1}{4}(x-4)

Step-by-step explanation:

since the equation is perpendicular to y=4x-2, m=-1/4 (negative reciprocal). (x_1,y_1)=(4,-11), plug all the values into the equation y- y_1 = m(x-x_1) and we get y+11=-\frac{1}{4}(x-4)

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Two angles of a triangle measures of 70° and 85° which is not the measure of an exterior angle of the triangle
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I hope this helps you

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