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Phantasy [73]
3 years ago
6

Please help its easy but i dont know it

Mathematics
1 answer:
IrinaK [193]3 years ago
7 0

Answer:

? = 12

Step-by-step explanation:

Find the average of the 2 end values, that is

? = \frac{26-2}{2} = \frac{24}{2} = 12

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SIMPLIFY 7/10+1/4-1/5=
Strike441 [17]

Answer:

3/4

Step-by-step explanation:

Q = \frac{7}{10} +\frac{1}{4} -\frac{1}{5}      

{the smallest common multiple of 10, 4 and 5 is 5×2×2 = 20 (denominator)}

Q =\frac{7×2}{10×2} +\frac{1×5}{4×5} -\frac{1×4}{5×4}

=\frac{14}{20} +\frac{5}{20} -\frac{4}{20}

=\frac{14+5-4}{20}

=\frac{15}{20}

=\frac{3×5}{4×5}

=\frac{3}{4}

4 0
4 years ago
Miss Silvia order 5 pizzas for every 20 students working on the campus clean up how many pizzas should I order if there are 36 s
Delvig [45]
12 for 48, 9 for 36.
3 0
3 years ago
84. Use properties of exponents to rewrite each expression with only positive, rational exponents. Then find the numerical value
Yakvenalex [24]

Answer:

a) 1/27

b) 16

c) 1/8

Step-by-step explanation:

a) x^{-3/2}

One of the properties of the exponents tells us that when we have a negative exponent we can express it in terms of its positive exponent by turning it into the denominator (and changing its sign), so we would have:

x^{-3/2}=\frac{1}{x^{3/2} }

And now, solving for x = 9 we have:

\frac{1}{x^{3/2}}=\frac{1}{9^{3/2} }  =\frac{1}{27}

b) y^{4/3}

This is already a positive rational exponent so we are just going to substitute the value of y = 8 into the expression

y^{4/3}=8^{4/3}=16

c) z^{-3/4}

Using the same property we used in a) we have:

z^{-3/4}=\frac{1}{z^{3/4} }

And now, solving for z = 16 we have:

\frac{1}{z^3/4} } =\frac{1}{16^{3/4} } =\frac{1}{8}

4 0
3 years ago
Plzzzzzzzzzzzzzzzzz help quick
Sergeeva-Olga [200]

Answer:

y=-2x-7 D

Step-by-step explanation:

3 0
4 years ago
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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