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3 years ago

If x represents a number then write an expression for a number that is eight less than twice the value of x.

Mathematics

2 answers:

Triss [41]3 years ago

5 0

2x-1

have a nice day

kherson [118]3 years ago

3 0

Twice the value of x is 2x, because it’s x with another x added on top (x + x = 2x)

Eight less than that is 2x - 8. That’s because you are taking away 8 from 2x (less than = subtraction).

Therefore your final expression is 2x - 8.

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if cot theta = -24/7 and cos theta < 0, name the quad in which theta lies, then find the exact value of csc theta

NemiM [27]

Cot θ = -24/7 and cos θ < 0

"cos θ < 0" means cosine is negative

Remember osmething like "ALL students Take Calculus"?

all is positive in quadrant 1
sine/cosecant is positive only in quadrant 2
tangent/cotangent is positive only in quadrant 3
cosine/secant is positive only in quadrant 4

since cot(angent) is the reciprocal of tangent, tangent must also be negative.

tangent is negative in quadrants II and IV
cosine is negative in quadrants II and III

therefore, θ must lie in quadrant II.

remember the tan θ = y / x. it is the y-coordinate divided by the x-coodinate. the y-coordinate would be the height of this 90 degree triangle that forms. the x-coordinate would be the bottom leg of the 90 degree triangle

then cot θ = x / y. therefore, cot θ = -24 / 7 means x = -24 and y = 7. we draw a triangle with these dimensions.

csc θ = r / y = hypotenuse over opposite (when you view the reference angle labeled as the one to base the hypotenuse/oppoiste on) because it is the reciprocal of sine

the hypotenuse can be found w/ pythagorean theorem

hypotenuse = √[(-24)² + 7²] = 25

therefore,
cot θ = 25/7

4 0

3 years ago

Fraction <img src="https://tex.z-dn.net/?f=%20-%206%20%5Cdiv%20%20%5Cfrac%7B4%7D%7B5%7D%20" id="TexFormula1" title=" - 6 \div

otez555 [7]

$-6\div\dfrac{4}{5}=-\dfrac{\not6^3}{1}\cdot\dfrac{5}{\not4_2}=-\dfrac{3}{1}\cdot\dfrac{5}{2}=-\dfrac{15}{2}=-7\dfrac{1}{2}$

8 0

3 years ago

Determine two coterminal angles (one positive and one negative) for the given angle. 52

Sphinxa [80]

-308° and 412° because 52-360 is -308 and 52+360 is 412

6 0

3 years ago

Compute the flux of curl(F) through the part of the paraboloid z = x 2 + y 2 that lies below the plane z = 4 with upward-pointin

kkurt [141]

Parameterize this surface (call it S) by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+u^2\,\mathbf k$

with $0\le u\le2$ and $0\le v\le2\pi$ .

The normal vector to S is

$\mathbf n=\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}=-2u^2\cos v\,\mathbf i-2u^2\sin v\,\mathbf j+u\,\mathbf k$

Compute the curl of F :

$\nabla\times\mathbf F=-2\,\mathbf i+3\,\mathbf j+5\,\mathbf k$

So the flux of curl(F) is

$\displaystyle\iint_S(\nabla\times\mathbf F)\cdot\mathrm d\mathbf S=\int_0^{2\pi}\int_0^2(\nabla\times\mathbf F)\cdot\mathbf n\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^2(5u+4u^2\cos v-6u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{20\pi}$

Alternatively, you can apply Stokes' theorem, which reduces the surface integral of the curl of F to the line integral of F along the intersection of the paraboloid with the plane z = 4. Parameterize this curve (call it C) by

$\mathbf r(t)=2\cos t\,\mathbf i+2\sin t\,\mathbf j+3\,\mathbf k$

with $0\le t\le2\pi$ . Then

$\displaystyle\iint_S(\nabla\times\mathbf F)\cdot\mathrm d\mathbf S=\int_0^{2\pi}\mathbf F\cdot\mathrm d\mathbf r$

$=\displaystyle\int_0^{2\pi}(20\cos^2t-24\sin t)\,\mathrm dt=\boxed{20\pi}$

8 0

4 years ago

E is an option btw but i need work provided plz!

enot [183]

PQ is a radius. QS is tangent to the circle. In this scenario (a radius and a tangent passing through the same point of the radius) the two segments are always perpendicular.

This implies that PQS is a right triangle. We know that PQ = 10 because it's a radius.

Also, we have

$PS=PR+RS = 10+16 = 26$

because PR is a radius and RS is given.

So, we can derive QS using the pythagorean theorem:

$QS = \sqrt{PS^2-PQ^2} = \sqrt{676-100} = \sqrt{576} = 24$

3 0

3 years ago