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2 years ago

Solving one - step equations

1 answer:

3 0

You will only need to perform one step in order to solve the equation. One goal in solving an equation is to have only variables on one side of the equal sign and numbers on the other side of the equal sign. The other goal is to have the number in front of the variable equal to one.

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What is two thirds or 2/3 of the whole 120

Zepler [3.9K]

It would be 80 because 2 times 120 is 240. 240 divided by 3 is 80

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2 years ago

Beth is putting liquid fertilizer on the plants in 4 flowerpots. her measuring spoon holds 1/8 teaspoon the directions say to pu

VARVARA [1.3K]

20 times 1/8+1/8+1/8+1/8+1/8=5/8 times 4 because of the four pots = 20

5 0

2 years ago

Read 2 more answers

Decreasing 40 to 20 is the same as decreasing 40 by one-half of . So, it is a 50% decrease.

Paha777 [63]

Yes it is also a 50% decrease.

7 0

2 years ago

If f(x) = 28+x + 7, what is the value of f(-9), to the nearest<br> hundredth (if necessary)?

GREYUIT [131]

Answer:

Step-by-step explanation:

Problem: f(x) = 28 + x + 7

Plug in f(-9)

New equation: f(-9) = 28 + (-9) + 7

Simplify: f(-9) = 26

<u>Answer: f(-9) = 26</u>

7 0

2 years ago

Read 2 more answers

Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x2 from (−1, 2) to (1, 2) (a) find a function f

Korvikt [17]

If there is some scalar function $f(x,y)$ such that

$\nabla f(x,y)=\mathbf f(x,y)=x^2\,\mathbf i+y^2\,\mathbf j$

then we want to find $f$ such that

$\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)$
$\dfrac{\partial f}{\partial y}=y^2=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\dfrac{y^3}3+C$
$\implies f(x,y)=\dfrac{x^3}3+\dfrac{y^3}3+C$

So the vector field $\mathbf f(x,y)$ is conservative, which means the fundamental theorem applies; the line integral of $\mathbf f$ along any path $\mathcal C$ parameterized by some vector-valued function $\mathbf r(t)$ over $a\le t\le b$ is given by

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=a}^{t=b}\mathbf f(\mathbf r(t))\cdot\dfrac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt=f(\mathbf r(b))-f(\mathbf r(a))$

In this case,

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(1,2)-f(-1,2)=\dfrac23$

5 0

2 years ago