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3 years ago

7

I really need help with this. It's on a test due today, and the teacher said we can use any help we can get, so this is allowed.

Thank you for whoever answers this.!

1 answer:

geniusboy [140]3 years ago

4 0

Answer:

I think it would be 88.

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What is 13/8 as a improper fraction

satela [25.4K]

13/8 is already an improper fraction

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3 years ago

Charlie is flying a kite one afternoon and steps on the end of the string to have his

Butoxors [25]

Answer:

roubd it to 20 and you will get 679.90

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3 years ago

BROOO HELP what is the slope of the line through (-3,3) and (-1,-1)?

lesya [120]

Answer:

C

Step-by-step explanation:

To find the slope between any two points, we can use the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

Where (x₁, y₁) and (x₂, y₂) are two, separate points.

We have the two points (-3, 3) and (-1, -1).

So, let (-3, 3) be (x₁, y₁) and let (-1, -1) be (x₂, y₂).

Substitute them into the slope formula to get:

$m=\frac{-1-3}{-1-(-3)}$

Subtract:

$m=\frac{-4}{2}=-2$

Hence, our slope is -2.

So, our answer is C.

6 0

3 years ago

Read 2 more answers

For each line, determine whether the slope is positive, negative, zero, or undefined.

strojnjashka [21]

Answer:

1. Undefined

2. Negative

3. Zero

4. Positive

Step-by-step explanation:

1. Undefined, the line moves in both directions vertically. It's impossible to tell what the slope is.

2. Negative, the line is decreasing at a constant rate, the slope is negative.

3. Zero, the line doesn't increase or decrease, therefore the slope is 0. (No changes)

4. Positive, the line is increasing at a large rate, the slope is positive.

5 0

3 years ago

Find the value or measure. Assume all lines that appear to be tangent are tangent. X=

aev [14]

According to the secant-tangent theorem, we have the following expression:

$(x+3)^2=10.8(19.2+10.8)$

Now, we solve for <em>x</em>.

$\begin{gathered} x^2+6x+9=10.8(30) \\ x^2+6x+9=324 \\ x^2+6x+9-324=0 \\ x^2+6x-315=0 \end{gathered}$

Then, we use the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Where a = 1, b = 6, and c = -315.

$\begin{gathered} x_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_{1,2}=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_{1,2}=\frac{-6\pm36}{2} \\ x_1=\frac{-6+36}{2}=\frac{30}{2}=15 \\ x_2=\frac{-6-36}{2}=\frac{-42}{2}=-21 \end{gathered}$ <h2>Hence, the answer is 15 because lengths can't be negative.</h2>

7 0

1 year ago