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2 years ago

Solve for x please... it's for a test and it's really hard

Mathematics

2 answers:

trasher [3.6K]2 years ago

6 0

X= -10 hope this helped

LUCKY_DIMON [66]2 years ago

3 0

The answer is x=-10 hope this is right

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Please help its hard btw

Rudiy27

Answer:

4213

Step-by-step explanation:

go to the numbers that pop out and do it backwards.

7 0

3 years ago

-0.4a + 3 = 7 a = plz help

Papessa [141]

Answer:

a = -10

Step-by-step explanation:

-0.4a + 3 = 7

subtract 3 from each side

-0.4a + 3-3 = 7-3

-.4a = 4

Divide each side by -.4

-.4a/ -.4 = 4/-.4

a = - 10

4 0

3 years ago

Elizabeth has $26 left after shopping at the mall. She bought 2 shirts for $22.99 each, a drink for $2.02, and 2 books for $16 e

ra1l [238]

Step-by-step explanation:

22.99 × 2 + 2.02 + 16 × 2 + 26= $106

3 0

3 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago

Question 1: Describe in words the region of double-struck R3 represented by the equation: x2 + z2 ≤ 36.

Ganezh [65]

Step-by-step explanation:

1) i will asume x2 and z2 are squares here.

ok, here we only have restrictions to x and z, so y can take all the values in R.

$x^{2} + z^{2} \leq 36$ is a circle of radius 6, you can see this if for example we set z = 0, then x goes from -6 to 6, the same if we set x = 0 then z goes from -6 to 6.

and the equation $x^{2} + z^{2} \\$ describes a circle.

So here, the region is the solid cylinder of radius 6, where the Y axis is also the axis of the cylinder.

2) you tipped the same inequality but different numbers in the right side, here i think you are saying that the inequalities describes the set of all points whose distance from the y-axis is equal or less tan 6.

7 0

3 years ago