All categories

4 years ago

-7=z/-6 solve for z pls help it 7th grade math and it’s on khan ;cc

Mathematics

2 answers:

trasher [3.6K]4 years ago

6 0

Answer:

Step-by-step explanation:

I think it's 42 because 42/-6 is -7

STatiana [176]4 years ago

5 0

Answer:

z = 42

Step-by-step explanation:

Given -7 = z/-6

Cross multiply

-7 x -6 = z

Negative times negative gives you positive.

42 = z

z = 42

You might be interested in

Geoffrey draws a triangle. One of the angles measures 75° and one of the angles measures 40°. What is the measure of the third a

valina [46]

Answer:

D) 65°

Step-by-step explanation:

Remember, all the angles of a triangle are 180 degrees and 75 + 40 + 65 = 180 degrees. Please if you can, support my channel. In the images you can find the information ;)

4 0

2 years ago

What is the correct slope and y-intercept for the following: y=-3x+8

Vlad [161]

━━━━━━━☆☆━━━━━━━

▹ Answer

Slope = -3

Y-intercept = 8

▹ Step-by-Step Explanation

y = mx + b

mx represents the slope.

b represents the y intercept.

therefore,

y = -3x + 8

Hope this helps!

CloutAnswers ❁

Brainliest is greatly appreciated!

━━━━━━━☆☆━━━━━━━

5 0

3 years ago

Read 2 more answers

Simplify the expression using the distributive property 42 + 7a

melamori03 [73]

42+7a= 7(6+a)
By noticing they both have a factor of 7, I was able to find the equation.

8 0

3 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$