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3 years ago

Help..!!! Why do i ask so many questions

Mathematics

1 answer:

guapka [62]3 years ago

5 0

Answer:

Hey there!

-4/3-4/5

-20/15-12/15

-32/15

Let me know if this helps :)

You might be interested in

Simplify the expression. Show all of your work. 16 plus 4 cubed times 2 divided by 16

frozen [14]

Answer:

528

Step-by-step explanation:

Question: 16 + $4^{3}$ * 2 ÷ 16

Work:

$4^{3}$ = 64

2/16 = 0.125

64/0.125 = 512

512 + 16 = 528

4 0

2 years ago

F(x)=3x/x-1 answer needed

Ray Of Light [21]

Depends on what your answering for, for this i’d assume ur answering for the function which the answer would be 3x/x-1 - f(x) = 0. let me know if it’s something else you are trying to solve for

5 0

3 years ago

If f(x)=4x^3 - 6x^2 + 2x - 5, then f(-2) = ___________

Art [367]

f(x)=4x^3 - 6x^2 + 2x - 5

f(-2)

Replace X with -2:

4(-2)^3 - 6(-2)^2 + 2(-2) - 5 =

Calulate the powers:

4(-8) - 6(4) +(-4) -5=

Multiply:

-32 - 24 -4 -5=

Subtract

-65

7 0

3 years ago

1. Find the remainder if f(x) = 2x³ + 8x² – 5x + 5 is divided by x – 2.

Hunter-Best [27]

Answer:

The answer to your question is 43

Step-by-step explanation:

2x³ + 8x² - 5x + 5 / x - 2

Process

1.- Use synthetic division

2 8 -5 5 2

4 24 38

2 12 19 43

Quotient 2x² + 12x + 19

Remainder 43

4 0

3 years ago

7x^2=9+x what are the values of x Will give medal and points

DENIUS [597]

7x² = 9 + x Subtract x from both sides
7x² - x = 9 Subtract 9 from both sides
7x² - x - 9 = 0 Use the Quadratic Formula

a = 7 , b = -1 , c = -9

x = $\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ Plug in the a, b, and c values
x = $\frac{- (-1) \pm \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}$ Cancel out the double negative
x = $\frac{1 \pm \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}$ Square -1
x = $\frac{1 \pm \sqrt{1 - 4(7)(-9)} }{2(7)}$ Multiply 7 and -9
x = $\frac{1 \pm \sqrt{1 - 4(-63} }{2(7)}$ Multiply -4 and -63
x = $\frac{1 \pm \sqrt{1 + 252} }{2(7)}$ Multiply 2 and 7
x = $\frac{1 \pm \sqrt{1 + 252} }{14}$ Add 1 and 252
x = $\frac{1 \pm \sqrt{253} }{14}$ Split up the $\pm$
x = $\left \{ {{ \frac{1 + \sqrt{253} }{14} } \atop { \frac{1 - \sqrt{253} }{14} }} \right.$
The approximate square root of 253 is 15.905973.
x ≈ $\left \{ { \frac{1 + 15.905973}{14} } \atop { \frac{1 - 15.905973}{14} }} \right$ Add and subtract
x ≈ $\left \{ {{ \frac{16.905973}{14} } \atop { \frac{14.905973}{14} }} \right.$ Divide
x ≈ $\left \{ {{1.2075} \atop {1.0647}} \right.$ Round to the nearest hundredth
x ≈ $\left \{ {{1.21} \atop {1.06}} \right.$

7 0

3 years ago