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myrzilka [38]
3 years ago
14

What is the length of the hypotenuse?

Mathematics
2 answers:
kotegsom [21]3 years ago
6 0

Answer:

10

Step-by-step explanation:

Pythagorean theorem

a^2 + b^2 = c^2

6^2 + 8 ^2 = c^2

36 + 48 = c^2

100 = c^2

10=c

siniylev [52]3 years ago
3 0

Answer:

90 degrees hope it helpz

Step-by-step explanation:

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What is the solution to the system of linear equations? x - 2y = -3
Diano4ka-milaya [45]
X= -7 and y= -2
Hope this helps
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8 0
3 years ago
4# is a two digit number ,where # represents the digit at ones place . 7999 ÷ 4# lies between ---------
Hitman42 [59]

Answer:

7999 \div 4\# lies between 163.245 and 199.975

Step-by-step explanation:

Given

2 digit = 4#

Required

The range of 7999 \div 4\#

Let

\# = 0 --- the smallest possible value of #

So:

7999 \div 40 = 199.975

Let

\# = 9 --- the largest possible value of #

So:

7999 \div 49 = 163.245

<em>Hence, </em>7999 \div 4\#<em> lies between 163.245 and 199.975</em>

4 0
3 years ago
A jar contains 4 green marbles and 3 blue marbles of the same size and shape. rachel will randomly pull 2 marbles from the bag a
vagabundo [1.1K]
Probably most likely 1.25% because in the question it doesn't say there is any blue marbles.
6 0
3 years ago
the diagram shows a rectangle is 13 cm the area of the rectangle is 93.6 cm calculate the width of the rectangle.
Korvikt [17]

Answer:

7.2 cm

Step-by-step explanation

The area of a rectangle is calculated by multiplying the width and the length together. Therefore:

13 multiplied by x = 93.6

x = 93.6 divided by 13

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3 0
3 years ago
Read 2 more answers
Find the volume of the solid under the plane 5x + 9y − z = 0 and above the region bounded by y = x and y = x4.
svp [43]
<span>For the plane, we have z = 5x + 9y

For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.

Since x > x^4 for y in [0, 1],

The volume of the solid equals

\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx  \\  \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx  \\  \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\  \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0

=\frac{19}{6} - \frac{5}{6} - \frac{1}{2} =\bold{ \frac{11}{6} \ cubic \ units}</span>
8 0
3 years ago
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