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4 years ago

Need problem answered plz!!!! 4^3⋅(100/25)−5^0

Mathematics

2 answers:

Llana [10]4 years ago

7 0

Use bodmas/pemdas rule

4^3*(4)-5^0
64*4-1 anything to the power zero is one
256-1
255

yarga [219]4 years ago

6 0

4^3 = 64
100/25=4
64x4=256
5^0=1
256-1=255
255

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48 divided by 76.8<br> I am very confused on the two digits numbers

Anestetic [448]

It’s 1.600 because it’a not the same when you round and it has different type of decimal places.

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2 years ago

Evaluate the polynomial function when H=3<br> f(h)=4h^3-3h^2+5h-16

tester [92]

Answer:

Step-by-step explanation:

f(3)=4(3)^3-3(3)^2+5(3)-16

f(3)=4(3*3*3)-3(3*3)+15-16

f(3)=4(27)-3(9)+15-16

f(3)=108-27+15-16

f(3)=81+15-16

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3 years ago

Solve the system by substitution.<br> 10x + 5y = -50<br> y = -x

Pie

Answer:

(1,-12)

Step-by-step explanation:

-10+5y=-50

5y=-60

y=-12

10x-60=-50

10x=10

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3 years ago

Complete the square to solve the equation below

frutty [35]

$x^2+x= \frac{7}{4}$

Take one-half of the coefficient of x and square it, then add it to both sides.

$x^2+x+ (\frac{1}{4}) =\frac{7}{4}+ (\frac{1}{4})$

$x^2+x+(\frac{1}{4})= 2$

$\left(x+\frac{1}{2}\right)^2=2$

$x+\frac{1}{2}=\pm \sqrt{2}$
-----------------------------------------------------------------
$x+\frac{1}{2}=- \sqrt{2}$
$x=-\frac{1}{2}- \sqrt{2}$ <=

AND

$x+\frac{1}{2}= \sqrt{2}$
$x= \sqrt{2}-\frac{1}{2}$ <=

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3 years ago

Solve the system of equations by elimination. Explain your reasoning or show your work.

saul85 [17]

9514 1404 393

Answer:

(x, y) = (5, 4)

Step-by-step explanation:

Subtracting the second equation from the first eliminates the x term.

(4x +6y) -(4x +2y) = (44) -(28)

4y = 16 . . . . simplify

y = 4

Using the second equation to find x, we have ...

4x +2(4) = 28

x + 2 = 7 . . . . . . divide by 4

x = 5 . . . . . . . . . subtract 2

The solution is (x, y) = (5, 4).

7 0

3 years ago