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3 years ago

Could you please help? I'm stuck on this.

Mathematics

1 answer:

Flauer [41]3 years ago

7 0

Answer:

Step-by-step explanation:

6 x 6=36

9 x 15=135

135-36=99

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Logarithmic differentiation: y=theta+8/theta *costheta

djyliett [7]

Y=Ф + 8/(ФcosФ) ==> note that

dy/dФ = 1 - 8/[(ФcosФ)]² x (-8sinФ) ==> dy/dФ = 1+[(64sinФ)/Фcos²Ф]

==> dy/dФ = 1+ 64tanФ/Ф or dy/dФ = (Ф+tanФ)/Ф

8 0

3 years ago

4(3xy^4)^3/(2x^3y^5)^4

myrzilka [38]

Answer:

$\frac{4\left(3xy^4\right)^3}{\left(2x^3y^5\right)^4}=\frac{27}{4x^9y^8}$

Step-by-step explanation:

Given the expression

$\:\:\frac{4\left(3xy^4\right)^3}{\left(2x^3y^5\right)^4}$

solving the expression

$\:\:\frac{4\left(3xy^4\right)^3}{\left(2x^3y^5\right)^4}=4\cdot \:\frac{\left(3xy^4\right)^3}{\left(2x^3y^5\right)^4}$

$=4\:\frac{27x^3y^{12}}{16x^{12}y^{20}}$

$=4\cdot \frac{3^3}{2^4x^9y^8}$

The multiply fractions are defined as

$\:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}$

so the expression becomes

$=\frac{3^3\cdot \:4}{2^4x^9y^8}$

$=\frac{3^3\cdot \:2^2}{2^4x^9y^8}$

$=\frac{3^3}{2^2x^9y^8}$

Refining

$=\frac{27}{4x^9y^8}$

Therefore,

$\frac{4\left(3xy^4\right)^3}{\left(2x^3y^5\right)^4}=\frac{27}{4x^9y^8}$

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3 years ago

The solution to which system of equations has an x-value of 3?

leva [86]

Answer:

Step-by-step explanation:

4 0

3 years ago

The line AB is shown on the grid.

Ilya [14]

Answer:

1) The gradient of a line perpendicular to AB.→ [-2}

2) The equation of the line perpendicular to AB and passing

through the midpoint of AB. →[y=-2x+6]

$---------$

hope it helps....

have a great day!!

3 0

3 years ago

Read 2 more answers

Solve the equation N/-2=-15

otez555 [7]

Answer:

Step-by-step explanation:

N/-2=-15

N= -15 times -2

N= 30

4 0

3 years ago

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