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2 years ago

What is the quotient of 1+ 3 Over 4 and 2 over 5

Mathematics

1 answer:

dolphi86 [110]2 years ago

5 0

Answer:

2 1/2

Step-by-step explanation:

(These are fractions)

1+3 ÷ 2

4 5

(Simplify)

4 ÷ 2

4 5

(Same denominator)

20 ÷ 8

20 20

(Divide)

2 1/2

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Please answer correctly !!!!!! Will mark brainliest !!!!!!!!!!!!!

romanna [79]

Answer:

x = - 9, x = 10

Step-by-step explanation:

Given

x² - x - 90 = 0

Consider the factors of the constant term (- 90) which sum to give the coefficient of the x- term (- 1)

The factors are - 10 and + 9, since

- 10 × 9 = - 90 and - 10 + 9 = - 1 , thus

(x - 10)(x + 9) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 10 = 0 ⇒ x = 10

x + 9 = 0 ⇒ x = - 9

lesser x = - 9

greater x = 10

5 0

2 years ago

I have nothing good to offer, but please help!

faltersainse [42]

Does the question have a photo attached on your screen because it sounds like the image you included is not completed

7 0

2 years ago

Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

case A) $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

case B) $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

case C) $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

case D) $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

the answer is

$f(x)=(x-2)(x+2)$

6 0

3 years ago

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A soft-drink machine is regulated so that the amount of drink dispensed is approximately normally distributed with standard devi

iVinArrow [24]

Answer:

Step-by-step explanation:

3 0

3 years ago

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How do you solve 10(z-2)=1+4

tamaranim1 [39]

I'm assuming you want us to solve for the unknown variable z

$10(z-2)=1+4$

Combine like terms on the right side

$10(z-2)=5$

Use the distributive property on the left side

$10z-20=5$

Add both sides by 20 to cancel out the "-20" on the left side

$10z=25$

Divide both sides by 10

$z=2.5$

That is the value of the known variable, z, in this equation. Let me know if you need any clarifications, thanks!

~ Padoru

6 0

2 years ago

Read 2 more answers