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Kaylis [27]
3 years ago
6

3/4 on a numberline

Mathematics
1 answer:
alexdok [17]3 years ago
4 0
This is where 3/4 would be on a number line(the one in red)

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Please help me! (´༎ຶོρ༎ຶོ`)
Pie

Answer: Honestly all the stuff we do in school will not be used in the real world

Step-by-step explanation: lowkey

6 0
3 years ago
Can you guys help me with my homework?
tatiyna

Answer:

3π=3×3.14 =9.42

7π=7×3.14=21.98

Step-by-step explanation:

so the distance around window , in feet be

3π=3×3.14=9.42

7π=7×3.14=21.98

8 0
2 years ago
Hello, happy Friday, I am just here with some geometry questions.
Hatshy [7]

Answer:

B

Step-by-step explanation:

So far, we know that:

∠D = ∠J.

And that:

DE:JK = 14:7 = 2:1

So, to prove that ΔDEF ~ ΔJKL by SAS, DF must be similar to JL, as those are the sides between the angle.  

So:

DF:JL = 2:1.

Our answer is B.

4 0
2 years ago
Which represents the solution set of 5(x+5) < 85?
Maru [420]

Answer:

\Huge \boxed{X

Step-by-step explanation:

To solve this problem, first you have to isolate it on one side of the equation. Remember to solve this problem, find represents the solution of 5(x+5)<85.

First, you divide by 5 from both sides.

\displaystyle \frac{5(x+5)}{5}

Solve.

\displaystyle 85\div5=17

\displaystyle x+5

Next, subtract 5 from both sides.

\displaystyle x+5-5

Solve.

\displaystyle 17-5=12

As a result, the correct answer is x<12.

<h2>Hope this helps! </h2><h2 /><h2>Have a wonderful blessing day! :)</h2><h2 /><h2>Good luck! :)</h2>
3 0
3 years ago
Explain why a quadratic equation with a positive discriminant has two real solutions, a quadratic equation with a negative discr
Bad White [126]

Answer:

A quadratic equation can be written as:

a*x^2 + b*x + c = 0.

where a, b and c are real numbers.

The solutions of this equation can be found by the equation:

x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}

Where the determinant is D = b^2 - 4*a*c.

Now, if D>0

we have the square root of a positive number, which will be equal to a real number.

√D = R

then the solutions are:

x = \frac{-b +- R }{2*a}

Where each sign of R is a different solution for the equation.

If D< 0, we have the square root of a negative number, then we have a complex component:

√D = i*R

x = \frac{-b +- C*i }{2*a}

We have two complex solutions.

If D = 0

√0 = 0

then:

x = \frac{-b +- 0}{2*a} = \frac{-b}{2a}

We have only one real solution (or two equal solutions, depending on how you see it)

3 0
3 years ago
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