Can someone help me with this math homework please!

Answer:

6 I think

Step-by-step explanation:

Answer:

200

Step-by-step explanation:

250/100=2.5

2.5*80=200

So the answer is 200

Answer:

soln,

set A ={ 1,2,3}

set B ={2,3}

Answer:

2x-y-1=0

Step-by-step explanation:

.

Answer:

Let that rational number to be subtracted be x.Given,-5/​6 - x = 4/9= - x = 4/9+5/6= - x = 23/18x = - 23/18.

Step-by-step explanation:

Answer:

TFY

Step-by-step explanation:

let's start with the 90 degrees angle.

this is C in the first, and T in the second triangle.

so, C and T must be aligned.

and the we go around.

F ~ H

and then

Y ~ S

Answer:

$765

Step-by-step explanation:

[tex]interest \: = \frac{prt}{100} \\ = \frac{(450)(5)(14)}{100} \\ = 315 \\ total \: money \: = 315 + 450 \\ = 765[/tex]

A(n) = 450 + (n – 1)(0.05 • 450); $765.00

Answer:-

The option B is the right answer.

Solution:-

[tex] \sf h = 2d = 2 \: • \: 2r = 4r[/tex]

[tex] \sf = \frac{1}{3} \pi {r}^{2} h[/tex]

[tex] \sf = \frac{1}{3} \pi {r}^{2} \: • \: 4r[/tex]

[tex] \sf = \frac{4}{3} \pi {r}^{3} \: \green✓ [/tex]

Answer: B

Step-by-step explanation:

To find the values of x, we first need to write the function into an equation. We can derive 2 equations from the problem.

Equation 1: y=2|x+6|-4

Equation 2: y=6

Now, we can substitute.

2|x+6|-4=6

Let's solve for x.

2|x+6|-4=6        [add both sides by 4]

2|x+6|=10          [divide both sides by 2]

|x+6|=5              [subtract both sides by 6]

x=-1

Now that we know x=-1 is one of the solutions, we can eliminate C and D.

We know that the absolute value makes the number inside positive always. Therefore, let's solve for x with -5 instead.

|x+6|=-5              [subtract both sides by 6]

x=-11

Therefore, we know that B is the correct answer.

=================================================

Explanation:

Consider something like 2b+6 factoring to 2(b+3). When we distribute that outer 2 back inside the parenthesis, we're multiplying that 2 by everything inside. Factoring goes in reverse of this and we divide each term of 2b+6 by the GCF 2.

The same thing applies to this current problem as well.

Divide each term by the -1.8 we want to factor out.

The results -2b and 5 will go inside the parenthesis. That's how we end up with -1.8(-2b+5)

You can use distribution to verify this

-1.8(-2b+5)

-1.8*(-2b) - 1.8*(5)

3.6b - 9

Answer:

[tex]y = |x+2|+1[/tex]

Step-by-step explanation:

Answer:

Okay

Step-by-step explanation:

Answer:

its to pixelated

Step-by-step explanation:

Answer:

8 cm and 9 cm

Step-by-step explanation:

Hi there!

The sum of the lengths of two sides of a triangle must always be greater than the length of the third side.

5 cm and 8 cm ⇒ 5+8=13; not greater than 13

6 cm and 7 cm ⇒ 6+7=13; not greater than 13

7 cm and 2 cm ⇒ 7+2=9; not greater than 13

8 cm and 9 cm ⇒ 8+9=17; greater than 13

Therefore, the last set of two sides is possible for the lengths of the the other two sides of this triangle.

I hope this helps!

Answer:

The answer is C 100πcm^3

9514 1404 393

Answer:

  (x, y) = (-2 12/17, -51)

Step-by-step explanation:

Here is the answer to ...

  [tex]4x+\dfrac{9}{y}+11=0\\\\\dfrac{6}{y}-3x=8[/tex]

If you mean something else, then parentheses are needed.

__

Let z = 1/y. Then the equations in general form are ...

  [tex]4x+9z+11=0\\3x-6z+8=0[/tex]

The solution (cross multiplication method) is ...

  x = (9·8 -(-6)·11)/(4(-6)-3·9) = 138/-51 = -2 12/17

  z = (11·3 -8·4)/-51 = -1/51

  y = 1/z = -51

The solution is (x, y) = (-2 12/17, -51).

Answer:

Step-by-step explanation:

as angles of two triangles are equal, so they are similar.

x/17 =35/14=40/16

x/17=35/14

x=35/14×17=85/2=42.5

perimeter of ΔJKL=40+35+42.5=117.5

A prime number has exactly two factors, 1 and itself. For example, 13 is a prime number because the only factors of 13 are 1 and 13. The number 8 is not prime because it has four factors: 1, 2, 4 and 8. The number 1 is not a prime number because it only has one factor (itself).

Answer:

1 / sqrt(3)

Step-by-step explanation:

tan(o) = sin(o) / cos(o)

sin(o) is the vertical distance from the x-axis. and that is in this basic circle the y-coordinate of the point.

cos(o) is the horizontal distance from the y-axis. that is the x-coordinate of the point.

so,

tan(o) = (1/2) / (sqrt(3)/2) = (2×1) / (2×sqrt(3)) = 1/sqrt(3)

Answer:

Step-by-step explanation:

Find the slope of QR. From that we can find the the slope of the line perpendicular to QR.

Q(-2, -5)   & R(8,1)

[tex]Slope \ = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{1-[-5]}{8-[-2]}\\\\=\frac{1+5}{8+2}\\\\=\frac{6}{10}\\\\=\frac{-3}{5}[/tex]

So, the slope of the line perpendicular to QR =  -1/m - 1÷ [tex]\frac{-5}{3} = -1*\frac{-3}{5}=\frac{3}{5}[/tex]

Bisector of QR divides the line QR to two half. We have find the midpoint of QR.

Midpoint = [tex](\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})[/tex]

               [tex]=(\frac{-2+8}{2},\frac{-5+1}{2})\\\\=(\frac{6}{2},\frac{-4}{2})\\\\=(3,-2)[/tex]

slope = 3/5  and the required line passes through (3 , -2)

y - y1 = m(x-x1)

[tex]y - [-2] = \frac{3}{5}(x - 3)\\\\y + 2 = \frac{3}{5}x-\frac{3}{5}*3\\\\y=\frac{3}{5}x-\frac{9}{5}-2\\\\y=\frac{3}{5}x-\frac{9}{5}-\frac{2*5}{1*5}\\\\y=\frac{3}{5}x-\frac{9}{5}-\frac{10}{5}\\\\y=\frac{3}{5}x-\frac{19}{5}[/tex]

Given :-

To Find :-

Solution :-

Taking the given equation ,

=> 7x -3 = 5x + 9

=> 7x - 5x = 9 + 3

=> 2x = 12

=> x = 12 ÷ 2

=> x = 6

Hence the required answer is 2 .

Answer:

Falso.

Step-by-step explanation:

Sea [tex]d = \frac{a}{b}[/tex] un número racional, donde [tex]a, b \in \mathbb{R}[/tex] y [tex]b \neq 0[/tex], su opuesto es un número real [tex]c = -\left(\frac{a}{b} \right)[/tex]. En el caso de elevarse a un exponente dado, hay que comprobar cinco casos:

(a) El exponente es cero.

(b) El exponente es un negativo impar.

(c) El exponente es un negativo par.

(d) El exponente es un positivo impar.

(e) El exponente es un positivo par.

(a) El exponente es cero:

Toda potencia elevada a la cero es igual a uno. En consecuencia, [tex]c = d = 1[/tex]. La proposición es verdadera.

(b) El exponente es un negativo impar:

Considérese las siguientes expresiones:

[tex]d' = d^{-n}[/tex] y [tex]c' = c^{-n}[/tex]

Al aplicar las definiciones anteriores y las operaciones del Álgebra de los números reales tenemos el siguiente desarrollo:

[tex]d' = \left(\frac{a}{b} \right)^{-n}[/tex] y [tex]c' = \left[-\left(\frac{a}{b} \right)\right]^{-n}[/tex]

[tex]d' = \left(\frac{a}{b} \right)^{(-1)\cdot n}[/tex] y [tex]c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{(-1)\cdot n}[/tex]

[tex]d' = \left[\left(\frac{a}{b} \right)^{-1}\right]^{n}[/tex]y [tex]c' = \left[(-1)^{-1}\cdot \left(\frac{a}{b} \right)^{-1}\right]^{n}[/tex]

[tex]d' = \left(\frac{b}{a} \right)^{n}[/tex] y [tex]c = (-1)^{n}\cdot \left(\frac{b}{a} \right)^{n}[/tex]

[tex]d' = \left(\frac{b}{a} \right)^{n}[/tex] y [tex]c' = \left[(-1)\cdot \left(\frac{b}{a} \right)\right]^{n}[/tex]

[tex]d' = \left(\frac{b}{a} \right)^{n}[/tex] y [tex]c' = \left[-\left(\frac{b}{a} \right)\right]^{n}[/tex]

Si [tex]n[/tex] es impar, entonces:

[tex]d' = \left(\frac{b}{a} \right)^{n}[/tex] y [tex]c' = - \left(\frac{b}{a} \right)^{n}[/tex]

Puesto que [tex]d' \neq c'[/tex], la proposición es falsa.

(c) El exponente es un negativo par.

Si [tex]n[/tex] es par, entonces:

[tex]d' = \left(\frac{b}{a} \right)^{n}[/tex] y [tex]c' = \left(\frac{b}{a} \right)^{n}[/tex]

Puesto que [tex]d' = c'[/tex], la proposición es verdadera.

(d) El exponente es un positivo impar.

Considérese las siguientes expresiones:

[tex]d' = d^{n}[/tex] y [tex]c' = c^{n}[/tex]

[tex]d' = \left(\frac{a}{b}\right)^{n}[/tex] y [tex]c' = \left[-\left(\frac{a}{b} \right)\right]^{n}[/tex]

[tex]d' = \left(\frac{a}{b} \right)^{n}[/tex] y [tex]c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{n}[/tex]

[tex]d' = \left(\frac{a}{b} \right)^{n}[/tex] y [tex]c' = (-1)^{n}\cdot \left(\frac{a}{b} \right)^{n}[/tex]

Si [tex]n[/tex] es impar, entonces:

[tex]d' = \left(\frac{a}{b} \right)^{n}[/tex] y [tex]c' = - \left(\frac{a}{b} \right)^{n}[/tex]

(e) El exponente es un positivo par.

Considérese las siguientes expresiones:

[tex]d' = \left(\frac{a}{b} \right)^{n}[/tex] y [tex]c' = \left(\frac{a}{b} \right)^{n}[/tex]

Si [tex]n[/tex] es par, entonces [tex]d' = c'[/tex] y la proposición es verdadera.

Por tanto, se concluye que es falso que toda potencia que se obtiene de elevar a un mismo exponente un número racional y su opuesto es la misma.