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Arlecino [84]
3 years ago
7

How do you solve this the 8 is for the whole base not just the right triangle just so you know

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
8 0
X = square root of 89
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What are the first two steps in solving the radical equation below?
Maru [420]

Answer:

B

Step-by-step explanation:

add 7 to get away the 7 on the left hand side

square both side to eliminate square root

6 0
2 years ago
I'm in algebra please help and explain with work
lidiya [134]
2x+4y=0
3x+y=10
So what we want to do is get y onto one side by itself using whatever equation you like.
So I'm taking 2x+4y=0 and subtracting 4y from both sides
So we end up with
2x=4y
Divide both sides by four
2/4x is 1/2 x
So we get 1/2x=y
Take this and plug it into the other equation
3x+y=10
3x+1/2x=10
Add your like terms
3 1/2x=10
Putting 3 1/2 as a mixed number we get 7/2
So..
7/2x=10
Multiply by the opposite of 7/2 on both sides
Which is 2/7 this will cancel out the left side.
So we end up with
X= 10•2/7
X=20/7
Which is 2 and 6/7 as a mixed number.

5 0
3 years ago
Alex is tracking how many computers he can fix in an afternoon. He listed the number of computers fixed per hour in the followin
Artist 52 [7]
Answer:
This is a linear function because there is a common difference of 4
⇒ 2nd answer
Step-by-step explanation:
- In the linear function there is a common difference between each two
consecutive data
- In the exponential function there is a common ratio between each two
consecutive data
- Lats check the data in the data in the table
(x) === 1 ⇒ 2 ⇒ 3
f(x) === 4 ⇒ 8 ⇒ 12
∵ x has consecutive numbers 1 , 2 , 3
∵ 8 - 4 = 4
∵ 12 - 8 = 4
∴ f(x) has a common difference 4
∵ 8 ÷ 4 = 2
∵ 12 ÷ 8 = 1.5
∴ f(x) has no common ratio
∴ The table represents a linear function with common difference 4
This is a linear function because there is a common difference of 4
6 0
3 years ago
IMPORTANT PLEASE!! 10 POINTS :)
s344n2d4d5 [400]

Answer:

The domain is all the possible inputs for this function. The range is all the possible outputs for this function.

The line starts at the <em>x-value</em> of 0 and ends at the <em>x-value</em> of 12. Therefore, your domain is D: [0, 12].

The line starts at the <em>y-value</em> of 9000 and ends at the <em>y-value</em> of 0. Therefore, your range is R: [0, 9000].

5 0
3 years ago
Please help with math and please show your work will give brainliest
Burka [1]

Answer:

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d,

if  and  then a=bc=da+c=b+d

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

3x+y=52x−y=0–––––––––––5x=5

The y’s add to zero and we have one equation with one variable.

Let’s try another one:

{x+4y=22x+5y=−2(5.3.3)

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.

This figure shows two equations. The first is negative 2 times x plus 4y in parentheses equals negative 2 times 2. The second is 2x + 5y = negative 2. This figure shows two equations. The first is negative 2x minus 8y = negative 4. The second is 2x + 5y = -negative 2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Add the equations yourself—the result should be −3y = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.

This figure shows two equations being added together. The first is negative 2x – 8y = −4 and 2x plus 5y = negative 2. The answer is negative 3y = negative 6.

We’ll do one more:

{4x−3y=103x+5y=−7

It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.

We can make the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12x and −12x.

This figure shows two equations. The first is 3 times 4x minus 3y in parentheses equals 3 times 10. The second is negative 4 times 3x plus 5y in parentheses equals negative 4 times negative 7.

This gives us these two new equations:

{12x−9y−12x−20y=30=28

When we add these equations,

\[{12x−9y=30−12x−20y=28–––––––––––––––––−29y=58

\]

the x’s are eliminated and we just have −29y = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution

Step-by-step explanation:

Here is some examples

4 0
2 years ago
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