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1 year ago

Which is a perfect square? 20,21, 24,25

Mathematics

2 answers:

Vesnalui [34]1 year ago

6 0

Answer:

20 is not a perfect square.

21 is not a perfect square.

24 is not a perfect square.

25 is a perfect square.

zubka84 [21]1 year ago

6 0

25 is a perfect square. 5 * 5 is 25

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4. Calculate the area of the triangle.<br> 20 mm<br> 11 mm

NARA [144]

I am assuming 20mm = breadth and 11mm = height

Ar of Triangle = 1/2 x b x h
= 1/2 x 20 x 11
= 110mm^2

6 0

10 months ago

Is the equation y=2x-2 is proportional or non-proportional?

Usimov [2.4K]

Answer:

i believe their proportional

Step-by-step explanation:

cause they have a linear function

8 0

2 years ago

20 POINTS AND BRAINLIEST FOR CORRECT ANSWER! A cone with height h and radius r has volume V = 13πr2h. If a certain cone with a h

lozanna [386]

Answer:

If a certain cone with a height of 9 inches has volume V = 3πx2 + 42πx + 147π, what is the cone’s radius r in terms of x?

Step-by-step explanation:

V = 3πx2 + 42πx + 147π

V=3π(x2 + 14x +49)

9.42(x2 + 14x +49)

9.42(x2 + 14x +14) -14 + 49= 0

9.42(x + 7)^2 + 35= 0

9.42(9.42(x + 7)^2 = - 35)9.42

(x + 7)^2 = - 35/9.42)

√(x + 7)^2=√- 35/9.42

x + 7 = - 1.927

x= - 1.927 - 7

x= - 8.927

V = 3π(- 8.927)^2 + 42π(- 8.927) + 147π

V=750.69 - 1177.29 + 461.58

h= 9 inches

V = 13πr2h

34.98 = 13(3.14) (r^2) (h)

34.98 = 40.82 (r^2) 9

34.98 = 367.38 r^2

34.98/ 367.38 = 367.38 r^2/ 367.38

0.095= r^2

3 0

2 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

The population of Corpus Christi Texas is 305,215 the population of Laredo Texas is 236,091 how many more people live in Corpus

Naddik [55]

Answer:

69,124

Step-by-step explanation:

subtract 305,215 by 236,091

305,215

- 236,091

5 0

2 years ago