All categories

2 years ago

Solve the following equation: (Find x) a) (x-5)²+(2x-10)(x+3)-x²+25

Mathematics

1 answer:

Alex73 [517]2 years ago

4 0

Answer:

hence the value of x is either 5 or 2

x=5 or x=2

You might be interested in

What is the answer to this solve for h 12h=180

Rzqust [24]

Divide 12 on both sides of the equation to get h by itself so it should be..15

7 0

3 years ago

Help calculus module 8 DBQ<br><br> please show work

igor_vitrenko [27]

1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of $R(t)$ at the endpoints of each subinterval. The sum is then

$\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}$

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since $R$ is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval $[a,b]$ , it guarantees the existence of some $c\in(a,b)$ such that

$\dfrac{R(b)-R(a)}{b-a)=R'(c)$

Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such $c$ for which $R'(c)=0$ . I would chalk this up to not having enough information.

3. $R(t)$ gives the rate of water flow, and $R(t)\approx W(t)$ , so that the average rate of water flow over [0, 8] is the average value of $W(t)$ , given by the integral

$R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt$

If doing this by hand, you can integrate by parts, setting

$u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt$

$\mathrm dv=\mathrm dt\implies v=t$

$R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)$

For the remaining integral, consider the trigonometric substitution $t=\sqrt 7\tan s$ , so that $\mathrm dt=\sqrt 7\sec^2s\,\mathrm ds$ . Then

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^{-1}\frac8{\sqrt7}\right)-\tan^{-1}\frac8{\sqrt7}\right)$

$\boxed{R_{\rm avg}=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^{-1}\frac8{\sqrt7}}$

or approximately 3.0904, measured in gallons per hour (because this is the average value of $R$ ).

4. By the fundamental theorem of calculus,

$g'(x)=f(x)$

and $g(x)$ is increasing whenever $g'(x)=f(x)>0$ . This happens over the interval (-2, 3), since $f(x)=3$ on [-2, 0), and $-x+3>0$ on [0, 3).

5. First, by additivity of the definite integral,

$\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt$

Over the interval [-2, 0), we have $f(x)=3$ , and over the interval [0, 6], $f(x)=-x+3$ . So the integral above is

$\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_{t=-2}^{t=0}+\left(-\dfrac{t^2}2+3t\right)\bigg|_{t=0}^{t=x}=\boxed{6+3x-\dfrac{x^2}2}$

6 0

3 years ago

What other information must be given in order to be able to prove the two triangles congruent by ASA?

Anastaziya [24]

The best and most correct answer among the choices provided by the question is the fourth choice. The information must be given in order to be able to prove the two triangles congruent by ASA would be <span>/PQ≅/QR. </span>I hope my answer has come to your help. God bless and have a nice day ahead!

4 0

3 years ago

an electronics company is designing a watch with a face that is in the shape of a hexagon and two congruent trapezoids attached.

nalin [4]

The total area of the face of the watch to the nearest tenth of a square centimemter is 9.0 cm²

Since an electronics company is designing a watch with a face that is in the shape of a hexagon and two congruent trapezoids attached. The heights of the trapezoids and the apothem of the hexagon measure 2 centimeters each, and the length of the shorter base of each trapezoid is 1.5 centimeters, the radii of the hexagon, and the base of the trapezoid form a triangle of

height, h = apothem of the hexagon = 2 cm and
base, b = length of shorter base of trapezoid.

<h3>Area of the triangle</h3>

So, the area of this triangle is A = 1/2bh

= 1/2 × 1. 5 cm × 2 cm

= 1.5 cm × 1 cm

= 1.5 cm²

<h3>Area of the hexagon</h3>

Since there are 6 of such triangles in the hexagon, the area of the hexagon, A' = 6A

= 6 × 1.5 cm²

= 9.0 cm²

So, the total area of the face of the watch to the nearest tenth of a square centimemter is 9.0 cm²

Learn more about area of a hexagon here:

brainly.com/question/369332

5 0

2 years ago

In a bag containing 29 marbles, 5 of the marbles are red and 2 are green. What is the probability of randomly selecting a marble

DedPeter [7]

Answer:

Probability= 22/29

Step-by-step explanation:

Total no. of marbles= 29

No of red marbles=5

No of green marbles= 2

No of marbles which are neither red nor green= 29-(5+2)

=29-7

=22 marbles

Probability= 22/29

7 0

4 years ago

Solve the following equation: (Find x) a) (x-5)²+(2x-10)(x+3)-x²+25​

Solve the following equation: (Find x) a) (x-5)²+(2x-10)(x+3)-x²+25