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3 years ago

1,034 / 95 as a mixed number

Mathematics

2 answers:

Goryan [66]3 years ago

7 0

I think it is 10 84/95

Ivanshal [37]3 years ago

6 0

Answer: 10 84/95

Step-by-step explanation:

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The area of a playground is 20 square yards. The length of the playground is 5 times longer than its width. Find the length and

jok3333 [9.3K]

The answer is <span>length = 10 yd, width = 2 yd
proof
</span><span>The length of the playground is 5 times longer than its width, it means
l=5w, but 10= 2x5 so we can write l= 5w,
and Area = 20 =l x w = (5x2) x 2 =20</span>

6 0

4 years ago

Juan bought a sweater for $15.95 and two shirts for $9.00 each. how much did juan spend on clothes?

icang [17]

Answer:

Step-by-step explanation:

Well first you have:

$15.95 for one sweater
$9.00 each for two shirts

You would double the amount $9.00 which would be:

9 + 9 = 18 or Multiply 9 x 2 = 18

So this would total to 15 + 18 = $33

6 0

2 years ago

-4x+y=6 -5x-y=21 solve by substiution

gulaghasi [49]

Answer:

Step-by-step explanation:

55 is your answer

8 0

3 years ago

Read 2 more answers

The center of an ellipse is (-9,3). one focus is (-7,3). The major axis is 24 units long. What is the equation of the ellipse in

CaHeK987 [17]

Answer:

$\frac{(x+9)^2}{144}+\frac{(y-3)^2}{140}=1$

Step-by-step explanation:

The standard equation of the ellipse is

$\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1\;\cdots(i)$

where, $(\alpha, \beta)$ is the center and a,b are semi axes of the ellipse along x-axis and y-axis respectively.

Given that, $(\alpha, \beta)=(-9,3)$ and major axis, $2a=24$ .

So, semi major axis, $a=24/2=12$ .

Now, focus of the ellipse is (-7,3).

Let $e$ be the eccentricity of the ellipse,

So, $e=\frac c a$

where $c$ is the distance between the center and focus of the ellipse.

By using the distance formula,

$c=\sqrt{(-9-(-7))^2+(3-3)^2}=2$

$\Rightarrow e=\frac {2}{12}=\frac{1}{6}\;\cdots(ii)$

Again, the relationship among $a,b$ and $e$ is

$e=\sqrt{1-\frac{b^2}{a^2}$

$\Rightarrow \frac{1}{6}=\sqrt{1-\frac{b^2}{(12)^2}$ [from equation (ii)]

$\Rightarrow \frac{1}{36}=\sqrt{1-\frac{b^2}{144}$ [squaring on both the sides]

$\Rightarrow \frac{b^2}{144}=1-\frac{1}{36}$

$\Rightarrow b^2=\frac{35}{36}\times144=140$

So, the value of square of semi-minir axis, $b^2=140$ .

Hence, from equation (i), the equation of required ellipse is standard form is

$\frac{(x-(-9))^2}{144}+\frac{(y-3)^2}{140}=1$

$\Rightarrow \frac{(x+9)^2}{144}+\frac{(y-3)^2}{140}=1$

8 0

3 years ago

Find the value of each variable.

insens350 [35]

Step-by-step explanation:

11)

x= 5

${y}^{2} = 2 \times {5}^{2} \\ {y}^{2} = 50 \\ y = 5 \sqrt{2}$

12)

$\tan(30 ) = \frac{9}{y} \\ y = \frac{9}{ \tan(30) } = 9 \sqrt{3} \\ {x}^{2} = {9}^{2} + {(9 \sqrt{3} )}^{2} \\ {x}^{2} = 324 \\ x = 18$

13)

$\sin(60) = \frac{x}{22} \\ x = 11 \sqrt{3} \\ \cos(60) = \frac{y}{22} \\ y = 11$

14)

${16}^{2} = {x}^{2} + {y}^{2} \\ x = y \\ {16}^{2} = 2 {x}^{2} \\ x = 8 \sqrt{2} \\ y = 8 \sqrt{2}$

15)

$\tan(30) = \frac{x}{42} \\ x = 14 \sqrt{3} \\ \cos(30) = \frac{42}{y} \\ y = \frac{42}{ \cos(30) } \\ y = 28 \sqrt{3}$

16)

${(19 \sqrt{2}) }^{2} = {x}^{2} + {y}^{2} \\ x = y \\ 722 = 2 {x}^{2} \\ {x}^{2} = 361 \\ x = 19 \\ y = 19$

4 0

3 years ago