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3 years ago

Find the real solution(s) by solving for x (round any decimals to the nearest 100th); 5x = 320

Mathematics

2 answers:

baherus [9]3 years ago

6 0

Answer:

x=64

Step-by-step explanation:

5x = 320

Divide each side by 5

5x/5 = 320/5

x=64

galben [10]3 years ago

5 0

The answer is x=64.

The picture attached is my work.

I basically divided both sides by 5 to get the x “alone”. The 5s cancel each other out. We are left with 320/5 which equals 64.

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Olivia is growing roses and keeps track of how much fertilizer (in ounces) she adds to the soil and how many blooms each rose bu

dusya [7]

Answer:

She needs to add no fertilizer, option A)

Step-by-step explanation:

Equation of a Line:

According to the conditions of the question, being x the ounces of fertilizer Olivia adds to the soil, and y the number of blooms each rose bush has, the relation between them is

$y = 1.345x + 4$

It's bee required to compute the value of x such that the roses have 4 blooms, thus

$1.345x + 4=4$

Solving for x

$1.345x=4-4=0$

$x=0$

This means she needs to add no fertilizer, option A)

8 0

3 years ago

Read 2 more answers

100 points!!! Pre calculus. I need helpppppppppp

Fantom [35]

Answer:

We have function,

$y = 3 - 6 \sin {}^{} (2x + \frac{\pi}{2} )$

Standard Form of Sinusoid is

$y = - 6 \sin(2x + \frac{\pi}{2} ) + 3$

Which corresponds to

$y = a \sin(b(x + c)) + d$

where a is the amplitude

2pi/b is the period

c is phase shift

d is vertical shift or midline.

In the equation equation, we must factor out 2 so we get

$y = - 6(2(x + \frac{\pi}{4} )) + 3$

Also remeber a and b is always positive

So now let answer the questions.

a. The period is

$\frac{2\pi}{ |b| }$

$\frac{2\pi}{ |2| } = \pi$

So the period is pi radians.

b. Amplitude is

$| - 6| = 6$

Amplitude is 6.

c. Domain of a sinusoid is all reals. Here that stays the same. Range of a sinusoid is [-a+c, a-c]. Put the least number first, and the greatest next.

So using that rule, our range is [6+3, -6+3]= [9,-3] So our range is [-3,9].

D. Plug in 0 for x.

$3 - 6 \sin((2(0) + \frac{\pi}{2} )$

$3 - 6 \sin( \frac{\pi}{2} )$

$3 - 6(1)$

$3 - 6$

$= - 3$

So the y intercept is (0,-3)

E. To find phase shift, set x-c=0 to solve for phase shift.

$x + \frac{\pi}{4} = 0$

$x = - \frac{\pi}{4}$

Negative means to the left, so the phase shift is pi/4 units to the left.

f. Period is PI, so use interval [0,2pi].

Look at the graph above,

6 0

2 years ago

ASAP!! Please help me. I will not accept nonsense answers, but will mark as BRAINLIEST if you answer is correctly with solutions

77julia77 [94]

Answer:

C. It has been stretched horizontally.

Step-by-step explanation:

According to the diagram, the parabola has not been reflected in any way, nor has it been translated, since the parabola is pointing up in a smile, and the vertex is still at the origin.

The parabola does appear to be stretched, since the parent function had points at (1, 1), and (2, 4), but this parabola does not contain those points.

Since the parabola appears to be flatter than the parent function, we can say that C. It has been stretched horizontally.

Hope this helps!

3 0

3 years ago

Read 2 more answers

In which quadrant is point (12,-12) located

Delvig [45]

It is located in quadrant 3. Hope this helped :D

5 0

3 years ago

Read 2 more answers

(PLEASE HELP ASAP)

UNO [17]

9514 1404 393

Answer:

A. 3×3

B. [0, 1, 5]

C. (rows, columns) = (# equations, # variables) for matrix A; vector x remains unchanged; vector b has a row for each equation.

Step-by-step explanation:

A. The matrix A has a row for each equation and a column for each variable. The entries in each column of a given row are the coefficients of the corresponding variable in the equation the row represents. If the variable is missing, its coefficient is zero.

This system of equations has 3 equations in 3 variables, so matrix A has dimensions ...

A dimensions = (rows, columns) = (# equations, # variables) = (3, 3)

Matrix A is 3×3.

B. The second row of A represents the second equation:

$0x_1+1x_2+5x_3=-1$

The coefficients of the variables are 0, 1, 5. These are the entries in row 2 of matrix A.

C. As stated in part A, the size of matrix A will match the number of equations and variables in the system. If the number of variables remains the same, the number of rows of A (and b) will reflect the number of equations. (The number of columns of A (and rows of x) will reflect the number of variables.)

6 0

2 years ago