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3 years ago

What is the equation of a line that is parallel to the line 2x + 5y = 10 and passes through the point (-5, 1)? Check all

Mathematics

2 answers:

Lilit [14]3 years ago

5 0

Answer:

B, E

Step-by-step explanation:

The equation of that line in slope intercept form is y = -2/5x + 2

The slope of a line parallel to that line will be the same as its slope, so -2/5.

To find the y-intercept of a line that passes through the point (-5, 1) with that slope, you will have to plug in the x and y values of that point into what you know of the equation of the line.

y = -2/5x + b

1 = -2/5(-5) + b

1 = 2 + b

-1 = b

From this, you can construct an equation.

y = -2/5x - 1

However, this is not an answer choice.

It cannot be A because this line does not have a slope of -1.

2x + 5y = 15

5y = -2x - 5

y = -2/5x - 1

It can be B because this line is the same as the equation we came up with.

It cannot be C because this line does not have a slope of -1 or a y-intercept of -3.

2x + 5y = -15

5y = -2x - 15

y = -2/5x - 3

It cannot be D because this line does not have a y-intercept of -3.

y - 1 = -2/5(x + 5)

y - 1 = -2/5x - 2

y = -2/5x - 1

It can be E because this equation matches the one we came up with.

vlabodo [156]3 years ago

5 0

Answer:

A: y = -2/5x - 1

B: 2x + 5y = -5

E: y - 1 = -2/5(x+5)

Step-by-step explanation:

I only got 2 out of 3 answers, but the ones listed above should be correct :) Hope this helps! :D

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Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial t

jekas [21]

Answer:

(12.1409, 14.0591

Step-by-step explanation:

Given that Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known.

Group Group One Group Two

Mean 87.600 74.500

SD 1.000 1.500

SEM 0.316 0.433

N 10 12

The mean of Group One minus Group Two equals 13.100

standard error of difference = 0.556

90% confidence interval of this difference:

$(13.1-1.725*0.556,13.1+1.725*0.556)\\=(12.1409, 14.0591)$

t = 23.5520

df = 20

4 0

3 years ago

Find the vertices and foci of the hyperbola. 9x2 − y2 − 36x − 4y + 23 = 0

Xelga [282]

Hey there, hope I can help!

NOTE: Look at the image/images for useful tips
$\left(h+c,\:k\right),\:\left(h-c,\:k\right)$

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1\:\mathrm{\:is\:the\:standard\:equation\:for\:a\:right-left\:facing:H}$
with the center of (h, k), semi-axis a and semi-conjugate - axis b.
NOTE: H = hyperbola

$9x^2-y^2-36x-4y+23=0 \ \textgreater \ \mathrm{Subtract\:}23\mathrm{\:from\:both\:sides}$
$9x^2-36x-4y-y^2=-23$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms}$
$9\left(x^2-4x\right)-\left(y^2+4y\right)=-23$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1$
$\frac{1}{1}\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$\frac{1}{1}\left(x^2-4x+4\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y+4\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)$

$\mathrm{Refine\:}-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right) \ \textgreater \ \frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=1 \ \textgreater \ Refine$
$\frac{\left(x-2\right)^2}{1}-\frac{\left(y+2\right)^2}{9}=1$

Now rewrite in hyperbola standardform
$\frac{\left(x-2\right)^2}{1^2}-\frac{\left(y-\left(-2\right)\right)^2}{3^2}=1$

$\mathrm{Therefore\:Hyperbola\:properties\:are:}\left(h,\:k\right)=\left(2,\:-2\right),\:a=1,\:b=3$
$\left(2+c,\:-2\right),\:\left(2-c,\:-2\right)$

Now we must compute c
$\sqrt{1^2+3^2} \ \textgreater \ \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \ 1^2 = 1 \ \textgreater \ \sqrt{1+3^2}$

$3^2 = 9 \ \textgreater \ \sqrt{1+9} \ \textgreater \ \sqrt{10}$

Therefore the hyperbola foci is at $\left(2+\sqrt{10},\:-2\right),\:\left(2-\sqrt{10},\:-2\right)$

For the vertices we have $\left(2+1,\:-2\right),\:\left(2-1,\:-2\right)$

Simply refine it
$\left(3,\:-2\right),\:\left(1,\:-2\right)$
Therefore the listed coordinates above are our vertices

Hope this helps!

8 0

3 years ago

A restaurant has a total of 30 tables which are of two

Anna007 [38]

Answer:

Total number of tables of first type = 23.

Total number of tables of second type = 7

Step-by-step explanation:

It is given that there are 30 tables in total and there are two types of tables.

Let's call the two seat tables, the first type as x and the second type as y.

∴ x + y = 30 ......(1)

Also a total number of 81 people are seated. Therefore, 2x number of people would be seated on the the first type and 5y on the second type. Hence the equation becomes:

2x + 5y = 81 .....(2)

To solve (1) & (2) Multiply (1) by 2 and subtract, we get:

y = 7

Substituting y = 7 in (1), we get x = 23.

∴ The number of tables of first kind = 23

The number of tables of second kind = 7

3 0

3 years ago

What is the equation in vertex form of a parabola with vertex (-2, -2) that opens downward and is stretched by a factor of 5?

Galina-37 [17]

Answer:

D). -5(x + 2)^2 - 2.

Step-by-step explanation:

Parabola which Opens down and vertex (-2, -2) is :

f(x) = a(x + 2)^2 - 2 where a is negative

Stretch by factor 5 gives:

-5(x + 2)^2 - 2.

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2 years ago

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Roman55 [17]

Answer:

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.005 = 1 liter

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have a good day, and be safe

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