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2 years ago

Isolate x in the first equation:

Mathematics

1 answer:

denis-greek [22]2 years ago

4 0

If y is substitute into either of the original equation. The solution as an ordered pair are: x=-2; y=3.

<h3>The solution as an ordered pair</h3>

x + 3y= 7 equation 1

x+2y=4 equation 2

Solving for y

x+3y-(x+2y)=7-4

x+3y-z-2y=7-4

3y-2y=7-4

y=7-4

y=3

Substitute y=3 into equation 2

x+2y=4

x+2×3=4

x+6=4

Collect like terms

x=4-6

x=-2

Therefore the solution as an ordered pair are: x=-2; y=3.

Learn more about solution as an ordered pair here:brainly.com/question/1515438

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2. Casey says, "I pick a number.

victus00 [196]

Step-by-step explanation:

7,14,21,28,35,42,49,56,63,70

18,36,54,72,90,108,126,144,162,180

4 0

3 years ago

Find the height of a rectangular prism if the surface are is 288 square centimetres and the base has a length of 4 centimetres a

liraira [26]

The rectangular prism below is made up of 6 rectangular faces, and the faces opposite to each other are have an equal area.
The area of one of the faces is lxh or lh, and the area of the face opposite to it is also equal, so the area of those two equal faces is 2lh.
The area of the base of the prism is lxw, and the area of its opposite face is equal, so the area of those two equal faces is 2lw.
The area of the face in the left side of the prism is wxh, and the area of the face opposite to it is also equal, so their combined area is 2wh.
So, to find the total surface area of a rectangular prism, we’d add the sum of all the rectangular faces, so the formula of finding the surface area of a rectangular prism=
2lh+2lw+2wh
=2(lh+lw+wh)
Given,
Surface area of the rectangular prism=288 cm^2
2(lh+lw+wh)=288
2[4h+(4x9)+9h)]=288
4h+(4x9)+9h=288/2
4h+(4x9)+9h=144
13h+36=144
13h=144-36
13h=108
h=108/13
h=8.308 cm
Hope this helps!

8 0

3 years ago

Kurt drives a total of 32 miles to and from work every day. He works 5 days a week. How many miles does Kurt drive in a week for

serg [7]

He drives 160 miles to and from work a week

4 0

3 years ago

Write each complex number in rectangular form. a. 2(cos(135)+i sin(135)) b. 3(cos(120))+i sin(120)) c. 5(cos(5pi/4)+i sin(5pi/4)

bagirrra123 [75]

$\bf r\left[ cos\left( \theta \right)+i\ sin\left( \theta \right) \right]\quad 
\begin{cases}
x=rcos(\theta )\\
y=rsin(\theta )
\end{cases}\implies 
\begin{array}{llll}
x&,&y\\
a&&b
\end{array}\implies a+bi\\\\
-------------------------------\\\\
2\left[ cos\left( 135^o\right)+i\ sin\left( 135^o\right) \right]\impliedby r=2\qquad \theta =135^o
\\\\\\
2\left( -\frac{\sqrt{2}}{2} \right)+i\ 2\left( \frac{\sqrt{2}}{2}\right)\implies -\sqrt{2}+\sqrt{2}\ i$

$\bf -------------------------------\\\\
3\left[ cos\left( 120^o\right)+i\ sin\left( 120^o\right) \right]\impliedby r=3\qquad \theta =120^o
\\\\\\
3\left( -\frac{1}{2} \right)+i\ 3\left( \frac{\sqrt{3}}{2}\right)\implies -\frac{3}{2}+\frac{3\sqrt{3}}{2}\ i$

$\bf \\\\
-------------------------------\\\\
5\left[ cos\left( \frac{5\pi }{4}\right)+i\ sin\left( \frac{5\pi }{4}\right) \right]\impliedby r=5\qquad \theta =\frac{5\pi }{4}
\\\\\\
5\left( -\frac{\sqrt{2}}{2} \right)+i\ 5\left( -\frac{\sqrt{2}}{2}\right)\implies -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\ i$

$\bf -------------------------------\\\\
4\left[ cos\left( \frac{5\pi }{3}\right)+i\ sin\left( \frac{5\pi }{3}\right) \right]\impliedby r=4\qquad \theta =\frac{5\pi }{3}
\\\\\\
4\left( \frac{1}{2} \right)+i\ 4\left( -\frac{\sqrt{3}}{2}\right)\implies \frac{1}{2}-\frac{\sqrt{3}}{2}\ i$

3 0

3 years ago

The number of errors in a textbook follow a poisson distribution with a mean of 0.01 errors per page. what is the probability th

neonofarm [45]

Mean number of errors in each page = 0.01
Mean number of errors in 100 pages = 0.01*100=1

It is possible to use the cumulative distribution function (CMF), but the math is a little more complex, involving the gamma-function. Tables and software are available for that purpose.

Thus it is easier to evaluate with a calculator for the individual cases of k=0,1,2 and 3.

The Poisson distribution has a PMF (probability mass function)

$P(k):=\frac{\lambda^ke^{-\lambda}}{k!}$
with λ = 1
=>
$P(0):=\frac{1^0e^{-1}}{0!}=0.3678794$
$P(1):=\frac{1^1e^{-1}}{1!}=0.3678794$
$P(2):=\frac{1^2e^{-1}}{2!}=0.1839397$
$P(3):=\frac{1^3e^{-1}}{3!}=0.0613132$
=>
$P(k$
or
P(k<=3)=0.9810 (to four decimal places)

3 0

2 years ago