All categories

3 years ago

Pls help due soon I just need help on how to solve problems like this

Mathematics

2 answers:

balu736 [363]3 years ago

4 0

Answer:

Ok bet

Step-by-step explanation:

3 to the power of four is 81

5 times 8 is 40

81+40=121

Always use pemdas

Illusion [34]3 years ago

4 0

Answer:

121

Step-by-step explanation:

$3^{4}$ =81

3×3×3×3

----------------

5×8 = 40

-----------------

81 + 40= 121

You might be interested in

This problem addresses some common algebraic errors. For the equalities stated below assume that x and y stand for real numbers.

weqwewe [10]

Answer:

The answers are:

a) $(x+y)^2 = x^2 + y^2$ is FALSE.

b) $(x+y)^2 = x^2 + 2xy + y^2$ is TRUE.

c) $\frac{x}{x+y}= \frac{1}{y}$ is FALSE.

d) $x-(x+y) = y$ is FALSE.

e) $\sqrt{x^2} = x$ is FALSE.

f) $\sqrt{x^2} = |x|$ is TRUE.

g) $\sqrt{x^2+4} = x+2$ is FALSE.

h) $\frac{1}{x+y} = \frac{1}{x} + \frac{1}{y}$ is FALSE.

Step-by-step explanation:

We can show that the FALSE statement are, in fact, false finding an example where the equality fails.

a) Take x=2 and y=3 and notice that (2+3)² = 25, while 2²+3²=4+9=13.

b) (x+y)² = (x+y)(x+y)=x²+xy+xy+y² = x² +2xy +y².

c) Take x=2 and y=3 and notice that 2/(2+3) = 2/5 which is different from 1/3.

d) Take x=2 and y=3 and notice that 2-(2+3) = 2-5=-3 which is different from 3. Recall that x-(x+y) = x-x-y=-y.

e) Take x=-2, and notice that [tax]\sqrt{(-2)^2} = \sqrt{4} = 2[/tex] which is different from -2.

f) The previous example illustrate why this is true.

g) Take x=1, and notice that $sqrt{1^2+4} = \sqrt{5}$ which is different from 3.

h) Take x=2 and y=3 and notice that 1/(2+3)=1/5 and 1/3+1/2 = 5/6.

3 0

3 years ago

at dinner you order a meal that costs $16.00 what is the total cost of the meal after a 20% tip and sales tax of 8%

Taya2010 [7]

20% = 3.20
8%=1.28
28%=4.48+16=20.48
The total cost would be $20.48

7 0

3 years ago

Read 2 more answers

Which of the following descriptions represent the transformation shown in the image? Part 1b

sveticcg [70]

Answer: b) 2 units left, 4 units up, reflect across y-axis

Step-by-step explanation:

The easiest way to do this is to graph the black triangle using the the given options to see which one lands on the red triangle.

The second option is the one that works.

4 0

3 years ago

there are 13 animals in a barn. some are ducks and some are pigs. there are 40 legs in all. how many of each animal are there?

konstantin123 [22]

Let X be chickens
Let Y be pigs,
then x+y=13
the animals have 2x+4y legs,
so 2x+4y=40
The system of the equations
[X+y=13
[2x+4y=40
Divide the second equation by 2
[x+y=13
[x+2y=20
subtract the first equation from the second one
y=20-13
y-7 pigs
substitute y=7 into the first equation
x+7=13
x=13-7
x=6 chickens

6 0

3 years ago

Solve using elimination x+y-2z=8 5x-3y+z=-6 -2x-y+4z=-13

Free_Kalibri [48]

So here is your answer with LaTeX issued format interpretation. Full process elucidated briefly, below:

$\begin{alignedat}{3}x + y - 2z = 8 \\ 5x - 3y + 2 = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}$

For this equation to get obtained under the impression of those variables we have to eliminate them individually for moving further and simplifying the linear equation with three variables along the axis.

Multiply the equation of x + y - 2z = 8 by a number with a value of 5; Here this becomes; 5x + 5y - 10z = 40; So:

$\begin{alignedat}{3}5x + 5y - 10z = 40 \\ 5x - 3y + z = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variable on our side, that is; "x":

5x - 3y + z = - 6

-

5x + 5y - 10z = 40
______________

- 8y + 11z = - 46

Therefore, we are getting.

$\begin{alignedat}{3}5x + 5y - 10z = 40 \\ - 8y + 11z = - 46 \\ - 2x - y + 4z = - 13 \end{alignedat}$

Multiply the equation of 5x + 5y - 10z = - 40 by a number with a value of 2; Here this becomes; 10x + 10y - 20z = 80.

Multiply the equation of - 2x - y + 4z = - 13 by a number with a value of 5; Here this becomes; - 10x - 5y + 20z = - 65; So:

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ - 10x - 5y + 20z = - 65 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variables on our side, that is; "x" and "z":

- 10x - 5y + 20z = - 65

+
10x + 10y - 20z = 80
__________________

5y = 15

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ 5y = 15 \end{alignedat}$

Multiply the equation of - 8y + 11z = - 46 by a number with a value of 5; Here this becomes; - 40y + 55z = - 230.

Multiply the equation of 5y = 15 by a number with a value of 8; Here this becomes; 40y = 120; So:

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 690 \\ 40y = 120 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variables on our side, that is; "y":

40y = 120

+

- 40y + 55z = - 230
_________________

55z = - 110

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 230 \\ 55z = - 110 \end{alignedat}$

Solving for the variable of 'z':

$\mathsf{55z = - 110}$

$\bf{\dfrac{55z}{55} = \dfrac{-110}{55}}$

Cancel out the common factor acquired on the numerator and denominator, that is, "55":

$z = - \dfrac{\overbrace{\sout{110}}^{2}}{\underbrace{\sout{55}}_{1}}$

$\boxed{\mathbf{z = - 2}}$

Solving for variable "y":

$\mathbf{\therefore \quad - 40y - 55 \big(- 2 \big) = - 230}$

$\mathbf{- 40y - 55 \times 2 = - 230}$

$\mathbf{- 40y - 110 = - 230}$

$\mathbf{- 40y - 110 + 110 = - 230 + 110}$

Adding the numbered value as 110 into this equation (in previous step).

$\mathbf{- 40y = - 120}$

Divide by - 40.

$\mathbf{\dfrac{- 40y}{- 40} = \dfrac{- 120}{- 40}}$

$\mathbf{y = \dfrac{- 120}{- 40}}$

$\boxed{\mathbf{y = 3}}$

Solve for variable "x":

$\mathbf{10x + 10y - 20z = 80}$

$\mathbf{Since, \: z = - 2; \quad y = 3}$

$\mathbf{10x + 10 \times 3 - 20 \times (- 2) = 80}$

$\mathbf{10x + 10 \times 3 + 20 \times 2 = 80}$

$\mathbf{10x + 30 + 20 \times 2 = 80}$

$\mathbf{10x + 30 + 40 = 80}$

$\mathbf{10x + 70 = 80}$

$\mathbf{10x + 70 - 70 = 80 - 70}$

$\mathbf{10x = 10}$

Divide by this numbered value $\mathbf{10}$ to get the final value for the variable "x".

$\mathbf{\dfrac{10x}{10} = \dfrac{10}{10}}$

The numbered values in the numerator and the denominator are the same, on both the sides. This will mean the "x" variable will be left on the left hand side and numbered values "10" will give a product of "1" after the division is done. On the right hand side the numbered values get divided to obtain the final solution for final system of equation for variable "x" as "1".

$\boxed{\mathbf{x = 1}}$

Final solutions for the respective variables in the form of " (x, y, z) " is:

$\boxed{\mathbf{\underline{\Bigg(1, \: \: 3, \: \: - 2 \Bigg)}}}$

Hope it helps.

8 0

3 years ago

Read 2 more answers