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3 years ago

Solve the system of equations.

Mathematics

1 answer:

expeople1 [14]3 years ago

8 0

Answer:(-5,-6)

Step-by-step explanation:

Solve for one variable then solve for the other. Don’t worry you will get a big fraction before you get to the whole number.

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Write an equation for a line parallel to y=-3x+2 that passes through the point (-1,4)

Wewaii [24]

y=−3x+2 is a linear equation in slope-intercept form with
slope m=(−3)
(and y-intercept =2)

We want the equation of a line with slope m=(−3) through the point (−2,−8)

Using the point-slope linear equation form:
(y−(−8))=(−3)(x−(−2))

Simplifying
y+8=−3x−6
or, in standard form
3x+y=−14

3 0

3 years ago

I need help please! The sequence 6,18,54,162, ... shows the number of pushups Kendall did each week, starting with her first wee

Romashka-Z-Leto [24]

Answer:

a: 3

b. 6973568802

Step-by-step explanation:

a₁ = 6 , r = 3 , a₂₀ =?

Result:

a₂₀ = 6973568802

Explanation:

To find a₂₀ we use the formula

aₙ = a₁ · r ^ⁿ⁻¹

In this example we have a₁ = 6 , r = 3 , n = 20. After substituting these values to above

formula, we obtain:

aₙ = a₁ · r ^ⁿ⁻¹

a₂₀ = 6 · 3 ^²⁰⁻¹

a₂₀ = 6 · 1162261467

a₂₀ = 6973568802

3 0

3 years ago

joe earns R25 per hour working at a pizza shop. How much will he earn over a weekend of two shifts, each six hours long

NeX [460]

Answer:

R300

Step-by-step explanation:

To solve:

Multiply 25 by 6 to represent the earnings of a shift.
Multiply the shift earnings by 2 to represent how much he earned over the weekend.

Multiply 25 by 6:

$25 \times 6 = 150$

Multiply 150 by 2:

$150 \times 2 = 300$

Joe will earn R300 over the weekend,

6 0

2 years ago

Which equation represents a linear function? Equation 1: y^3 = 2x +1 Equation 2: y= 3x +1 Equation 3: y= 5x^2 - 1 Equation 4: y=

Yuri [45]

the second equation is linear

8 0

3 years ago

Find the derivative of the function at P 0 in the direction of A. f(x,y,z) = 3 e^x cos(yz), P0 (0, 0, 0), A = - i + 2 j + 3k

Alik [6]

The derivative of $f(x,y,z)$ at a point $p_0=(x_0,y_0,z_0)$ in the direction of a vector $\vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k$ is

$\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}$

We have

$f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k$

and

$\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}$

Then the derivative at $p_0$ in the direction of $\vec a$ is

$3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}$

3 0

3 years ago