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3 years ago

8

Which of the following would be the correct form of the equation 10t² - 29t = -10 to be able to solve using the zero product pro

perty?

1 answer:

timofeeve [1]3 years ago

7 0

the correct answer is

B

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Loook at the image bellow <3<br> ill also give brainliest

Serhud [2]

Answer:

kinda late but the answer is B

Hope this helps!! <3

8 0

2 years ago

-4x+2=42how do I solve this problem

jeka94

Answer:

x = -10

Step-by-step explanation:

move all the constants (numbers without variables) to one side to get

-4x = 42 - 2

-4x = 40

multiply by -1

4x = -40

divide by 4

x = -10

3 0

2 years ago

Read 2 more answers

What is the total area of a cylinder that has a radius of 5 cm and a height of 12 cm

Grace [21]

376.99 try that the formula is 2x3.14x5x12

5 0

3 years ago

What expression shows the relationship between the value of any term and n, its position in the sequence for the given sequence?

san4es73 [151]

Answer:

a(n)=3+(n-1)2

Step-by-step explanation:

a(n)=3+(n-1)2

arithmetic sequence

5 0

2 years ago

Solve the following system of equations. Write each of your answers as a fraction reduced to lowest terms. In other words, write

morpeh [17]

Answer:

x = 57/28

y = -95/84

z = 97/168

Step-by-step explanation:

Use the application in the next link: https://www.zweigmedia.com/RealWorld/tutorialsf1/scriptpivotold.html

Start with the expanded array:

$\left[\begin{array}{cccc}1&5&8&1\\3&2&2&5\\-2&-7&2&5\\\end{array}\right]$

then using the tool provided, make row operations until you find the solution:

r2 = r2-3r1

$\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\-2&-7&2&5\\\end{array}\right]$

r3 = r3+2r1

$\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\0&3&18&7\\\end{array}\right]$

r2 = r2*(-1/13)

$\left[\begin{array}{cccc}1&5&8&1\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]$

r1 = r1- r2*5

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]$

r3 = r3+ r2*-3

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&168/13&97/13\\\end{array}\right]$

r3 = r3*13/168

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&1&97/168\\\end{array}\right]$

r2 = r2- r3*22/13

$\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]$

r2 = r2+ r3*6/13

$\left[\begin{array}{cccc}1&0&0&57/28\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]$

Here you have a reduced array an therefore the answers to each variable are on each row:

$\left[\begin{array}{c}x\\y\\z\end{array}\right]$

8 0

2 years ago