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Svetlanka [38]
2 years ago
5

Find the equation of the line that contains The given point and is perpendicular to the given line. Write the equation in slope

intercept form, if possible. (21,7); 7x-5y=6
​
Mathematics
1 answer:
mixer [17]2 years ago
8 0

Answer:

an = 63(-1/3)^(n-1)

Step-by-step explanation:

This is a geometric sequence with first term 63 and common ratio -1/3.

The equation for the nth term  is

an = 63(-1/3)^(n-1).

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Share £200 between Alice and Brian in the ratio 3 : 5<br>​
Pepsi [2]

Answer:

75:125=3:5

Step-by-step explanation:

1. Add the total number of parts

3+5=8

2. Find out how much one part is

200/8=25

3. Multiply each ratio by 25

3*25=75

5*25=125

4. Check they add up to 200

75+125=200

8 0
2 years ago
Read 2 more answers
ASAP Which power can you write to represent the volume of the cube shown? Enter the power as an expression with a base and an ex
n200080 [17]

Answer:

The power for the volume of the cube is '3'.

Exponent form of the volume is

Volume of the cube is  inch³.

Step-by-step explanation:

We are given,

A cube with side equal to  inches.

As we know,

Volume of a cube =

Thus, we get,

Volume of the given cube is,

Volume =

i.e. Volume =  inch³

Thus, we have,

The power for the volume of the cube is '3'.

Exponent form of the volume is

Volume of the cube is  inch³.

5 0
2 years ago
Read 2 more answers
Really need help on this asap!!!!!
marusya05 [52]

Answer:

srry

Step-by-step explanation:

6 0
3 years ago
X^2 -25 =0. Factor and use the zero product property to solve.
Gwar [14]
X^2 - 25 is a difference of squares which has a special factorization.

In general, a^2 - b^2 = (a + b)(a - b)

x^2 - 25 = 0

(x + 5)(x - 5) = 0

x + 5 = 0  or   x - 5 = 0

x = -5   or   x = 5
7 0
3 years ago
Read 2 more answers
Section 5.2 Problem 17:
Elina [12.6K]

This DE has characteristic equation

4r^2 - 12r + 9r = (2r - 3)^2 = 0

with a repeated root at r = 3/2. Then the characteristic solution is

y_c = C_1 e^{\frac32 x} + C_2 x e^{\frac32 x}

which has derivative

{y_c}' = \dfrac{3C_1}2 e^{\frac32 x} + \dfrac{3C_2}2 x e^{\frac32x} + C_2 e^{\frac32 x}

Use the given initial conditions to solve for the constants:

y(0) = 3 \implies 3 = C_1

y'(0) = \dfrac52 \implies \dfrac52 = \dfrac{3C_1}2 + C_2 \implies C_2 = -2

and so the particular solution to the IVP is

\boxed{y(x) = 3 e^{\frac32 x} - 2 x e^{\frac32 x}}

8 0
2 years ago
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