How do you find the LCM???

Find the third side in simplest radical form: 25

Gre4nikov [31]

Answer:

<h3> $\boxed{ \bold{24}}$ </h3>

Step-by-step explanation:

$\mathsf{given}$

$\mathsf{hypotenuse(h) = 25}$

$\sf{perpendicular (p) = 7}$

$\sf{base(b) = }$ ?

Now, Using Pythagoras theorem

$\sf{{h}^{2} = {p}^{2} + {b}^{2} }$

plug the values

⇒ $\sf{ {25}^{2} = {7}^{2} + {b}^{2} }$

Evaluate the power

⇒ $\sf{625 = 49 + {b}^{2} }$

Swap the sides of the equation

⇒ $\sf{49 + {b}^{2} = 625}$

Move constant to right hand side and change it's sign

⇒ $\sf{ {b}^{2} = 625 - 49}$

Calculate the difference

⇒ $\sf{ {b}^{2} = 576}$

Squaring on both sides

⇒ $\sf{b = 24}$

Hope I helped!

Best regards!

8 0

3 years ago

Item 1

AleksandrR [38]

It took them 4 days to install 480 chairs which means if they were working at a constant rate they got 120 chairs installed daily. with 360 chairs left that means in 3 days they will finish. so overall, it took the workers 7 days to finish, or a week.

8 0

3 years ago

Describe the transformation that would Change f(x) in the following ways: 1. Vertical reflection across the x axis 2. Vertical s

ElenaW [278]

You have to add the same number to keep

3 0

2 years ago

90% of what number is 57

aliina [53]

To solve this, set up the equation, 90% of x being .9*x=57. Then, solve for x, x=57/.9= $6 \frac{1}{3}$ =6.333

4 0

3 years ago

How do you solve System of Equations?

Katarina [22]

The addition method of solving systems of equations is also called the method of elimination. This method is similar to the method you probably learned for solving simple equations.

If you had the equation "x + 6 = 11", you would write "–6" under either side of the equation, and then you'd "add down" to get "x = 5" as the solution.

x + 6 = 11
–6 –6
x = 5

You'll do something similar with the addition method.

Solve the following system using addition.2x + y = 9
3x – y = 16Note that, if I add down, the y's will cancel out. So I'll draw an "equals" bar under the system, and add down:2x + y = 9
3x – y = 16
5x = 25Now I can divide through to solve for x = 5, and then back-solve, using either of the original equations, to find the value of y. The first equation has smaller numbers, so I'll back-solve in that one:2(5) + y = 9
 10 + y = 9
 y = –1Then the solution is (x, y) = (5, –1).

It doesn't matter which equation you use for the backsolving; you'll get the same answer either way. If I'd used the second equation, I'd have gotten:

3(5) – y = 16
 15 – y = 16
 –y = 1
 y = –1

...which is the same result as before.

Solve the following system using addition.x – 2y = –9
x + 3y = 16Note that the x-terms would cancel out if only they'd had opposite signs. I can create this cancellation by multiplying either one of the equations by –1, and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the –1 through the entire equation. (That means both sides of the "equals" sign!)I'll multiply the second equation.The "–1R2" notation over the arrow indicates that I multiplied row 2 by –1. Now I can solve the equation "–5y = –25" to get y = 5. Back-solving in the first equation, I get:x – 2(5) = –9
x – 10 = –9
x = 1Then the solution is (x, y) = (1, 5).

A very common temptation is to write the solution in the form "(first number I found, second number I found)". Sometimes, though, as in this case, you find the y-value first and then the x-value second, and of course in points the x-value comes first. So just be careful to write the coordinates for your solutions correctly. Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved

Solve the following system using addition.2x – y = 9
3x + 4y = –14Nothing cancels here, but I can multiply to create a cancellation. I can multiply the first equation by 4, and this will set up the y-terms to cancel.Solving this, I get that x = 2. I'll use the first equation for backsolving, because the coefficients are smaller.2(2) – y = 9
4 – y = 9
–y = 5
y = –5The solution is (x, y) = (2, –5). Solve the following system using addition. 
 4x – 3y = 25
–3x + 8y = 10Hmm... nothing cancels. But I can multiply to create a cancellation. In this case, neither variable is the obvious choice for cancellation. I can multiply to convert the x-terms to 12x's or the y-terms to 24y's. Since I'm lazy and 12 is smaller than 24, I'll multiply to cancel the x-terms. (I would get the same answer in the end if I set up the y-terms to cancel. It's not that how I'm doing it is "the right way"; it was just my choice. You could make a different choice, and that would be just as correct.)I will multiply the first row by 3 and the second row by 4; then I'll add down and solve.
Solving, I get that y = 5. Neither equation looks particularly better than the other for back-solving, so I'll flip a coin and use the first equation.4x – 3(5) = 25
4x – 15 = 25
4x = 40
x = 10Remembering to put the x-coordinate first in the solution, I get:(x, y) = (10, 5)

Usually when you are solving "by addition", you will need to create the cancellation. Warning: The most common mistake is to forget to multiply all the way through the equation, multiplying on both sides of the "equals" sign. Be careful of this.

Solve the following using addition.12x – 13y = 2
–6x + 6.5y = –2I think I'll multiply the second equation by 2; this will at least get rid of the decimal place.Oops! This result isn't true! So this is an inconsistent system (two parallel lines) with no solution (with no intersection point).no solution Solve the following using addition.12x – 3y = 6
4x – y = 2I think it'll be simplest to cancel off the y-terms, so I'll multiply the second row by –3.Well, yes, but...? I already knew that zero equals zero. So this is a dependent system, and, solving for "y =", the solution is:y = 4x – 2

(Your text may format the answer as "(s, 4s – 2)", or something like that.)

6 0

4 years ago

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