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4 years ago

Subtract the following polynomials. 3.1x + 2.8z 4.3x - 1.2z PLEASE HELP

Mathematics

1 answer:

Assoli18 [71]4 years ago

7 0

Answer:

3.1x+2.8z

3.1 over 10x +14 over 15 z

1 over 10 x (3.1x + 2.8z)

4.3x - 1.2z =

43 over 10x - 6 over 5z

1 over 10 x (43x-12z)

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Solve for this equation z^2-12z=-13

ASHA 777 [7]

Answer:

$z=\sqrt{-1}$

Step-by-step explanation:

$z^2-12z=-13\\z^2-\frac{-12}{+12} =-13+12\\z^2=-1\\\sqrt{z^{2} } =\sqrt{-1}\\ z=\sqrt{-1}$

5 0

3 years ago

Gwen volunteered to work at the ticket booth for her school's Halloween carnival. The chart below gives the number of hours Gwen

n200080 [17]

A linear relation is a relationship between two variables that when the corresponding pair of values of the variables are plotted, a straight line is obtained

The equation for the relationship between hours Gwen worked and number of tickets is t = 15·h

The reason the above equation is correct is as follows:

The given parameters are;

The table of values;

Hours(h): 1, 2, 3, 4

Tickets (t): 15, 30, 45, 60

Required: To write an equation that gives the relation between the number of hours Gwen worked and the number of tickets sold

Solution:

From the given table, we have;

The common difference between successive h-values is 1

The common difference between successive t-values is 30 - 15 = 45 - 30 = 60 - 45 = 15 = Constant

Given that the common difference of the h, and t-values are constant, the relationship is a linear relationship of the form, t = m·h + c

Where;

m = The rate of change of t with h, as follows;

m = (30 - 15)/(2 - 1) = 15

∴ t - 15 = 15 × (h - 1)

t - 15 = 15·h - 15

t = 15·h - 15 + 15 = 15·h

t = 15·h

The equation for the relation between the number of hours Gwen worked and the number of tickets she sold is t = 15·h

Learn more about mathematical relationships here:

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7 0

2 years ago

3(a+b)=19 if a=6, then a+b=?

aniked [119]

Hello!

In a really simple way, 3 times our answer is 19. We can just divide to find it.

19/3= 6 1/3

Anyway, it might be nice to solve the equation for b. If a=6, then b has to be 1/3, but let's check.

18+3b=19

3b=1

b=1/3

Therefore, a+b= 6 1/3

I hope this helps!

6 0

3 years ago

Read 2 more answers

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$

3 0

2 years ago

A seamstress has 1/4 of a foot of thread.Then she uses 1/5 of a foot of thread to sew a hole How much of the thread is left?

Whitepunk [10]

1/10 of the thread is left because a connected to b is the thread and minus the 1/5 is 1/10 since 1/3 +total thread to begin wit.

3 0

3 years ago

Read 2 more answers