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3 years ago

What decimal point is between 1.5 and 1.7

Mathematics

1 answer:

tangare [24]3 years ago

3 0

1.6 is between 1.7 and 1.5 because you see how the number is between the number

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What is the value of cos theta given that (- 5, - 4) is a point on the terminal side of theta ?

Arte-miy333 [17]

Answer:I believe it's -5√41/41

Step-by-step explanation:

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3 years ago

An X-ray machine for animals costs$125,000. If the loan is for 10 years, an the interest paid is $65,625, what is the interest r

Rina8888 [55]

The answer is 5.25%.

$65,625/10yrs=$6,562.50 per a year.

6,562.5/125000= .0525 which translates to 5.25%.

3 0

3 years ago

Please help!!!! I’ll mark you as brainliest if correct

vampirchik [111]

Answer:

You are correct

Step-by-step explanation:

The mode is the value that occurs most often is a data set

The median is the value in the middle

Mean is the average of all the points

Range is the largest minus the smallest

8 0

3 years ago

Read 2 more answers

Plss solve this problem... will mark brainiest

Sonja [21]

Answer:

87.97 or 88

Step-by-step explanation:

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3 years ago

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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

3 years ago