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3 years ago

Find the value of each variable. Write an equation then solve showing ALL the work A2x+10°

Mathematics

1 answer:

AveGali [126]3 years ago

4 0

Answer:

x = 25

Step-by-step explanation:

The three straight lines in the sides of the triangle means that ALL 3 SIDES ARE EQUAL.

This is an equilateral triangle. So all 3 angles are equal.

We know sum of 3 angles in a triangle is 180. Since 3 of the angles are equal, each angle is:

180/3 = 60

The top angle is given as "2x + 10", thus we can say "2x + 10" is equal to 60 degrees. Now we equate and solve for x:

$2x+10=60\\2x=60-10\\2x=50\\x=\frac{50}{2}\\x=25$

The value of x is 25

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A horizontal line passes through the point (5,-1). Which point is also on this line?

kramer

Answer:

(4,-1), (3,-1),(2,-1) or (1,-1)

if you draw a horizontal line through the point (5,-1) you will notice it crosses through every point ending with -1

3 0

4 years ago

The equation of a circle is x2 + y2 + 6x + 4y + 10 = 1. What is this equation written in its standard form?

Marrrta [24]

Option A: $(x+3)^2+(y+2)^2=4$ is the equation of the circle

Explanation:

The equation of the circle is $x^{2} +y^2+6x+4y+10=1$

We need to write the equation of the circle in standard form.

Let us subtract 10 from both sides of the equation.

Thus, we have,

$x^{2} +y^2+6x+4y=-9$

Grouping the terms,we have,

$(x^{2} +6x)+(y^2+4y)=-9$

Now, we shall complete the square, let us add and subtract the equation by 9 to write the term $\left(x^{2}+6 x\right)$ in the form of $(a+b)^2$

Thus, we have,

$(x^{2} +6x+9-9)+(y^2+4y)=-9$

Simplifying, we get,

$(x^{2} +6x+9)+(y^2+4y)=-9+9$

$(x+3)^2+(y^2+4y)=0$

Similarly, we shall complete the square for the term $\left(y^{2}+4 y\right)$ in the form of $(a+b)^2$

Thus, we have,

$(x+3)^2+(y^2+4y+4-4)=0$

Simplifying, we get,

$(x+3)^2+(y^2+4y+4)=4$

$(x+3)^2+(y+2)^2=4$

Thus, the equation of the circle in standard form is $(x+3)^2+(y+2)^2=4$

Hence, Option A is the correct answer.

3 0

4 years ago

A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and

allsm [11]

(a) Let $A(t)$ denote the amount of sugar in the tank at time $t$ . The tank starts with only pure water, so $\boxed{A(0)=0}$ .

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(<em>A(t)</em>/1080 kg/L) * (7 L/min) = 7<em>A(t)</em>/1080 kg/min

so that the net rate of change of $A(t)$ is governed by the ODE,

$\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}$

$A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}$

Multiply both sides by the integrating factor $e^{7t/1080}$ to condense the left side into the derivative of a product:

$e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}$

$\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}$

Integrate both sides:

$e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt$

$e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C$

Solve for $A(t)$ :

$A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}$

Given that $A(0)=0$ , we find

$0=\dfrac{378}5+C\implies C=-\dfrac{378}5$

so that the amount of sugar at any time $t$ is

$\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}$

(c) As $t\to\infty$ , the exponential term converges to 0 and we're left with

$\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5$

or 75.6 kg of sugar.

7 0

3 years ago

Use the whole number product and place value reasoning to place the decimal point in the second product. Explain how you know.

BARSIC [14]

The answer to ypur question is 48.24

5 0

4 years ago

Read 2 more answers

Suppose a movie starts at 5:00 p.m. and Lindsay, a customer who is always late, arrives at the movie theater at a random time be

Arte-miy333 [17]

Answer:

a) We know that the distribution for X is given by: $X \sim Unif (a=10, b=45)$

The density function for this distirbution is given by:

$f(x) = \frac{1}{b-a}= \frac{1}{45-10}= \frac{1}{35} , 10 \leq X \leq 45$

So then the height of the uniform density curve would be given by:

$h = \frac{1}{35}= 0.029$

b) For this case we want to calculate the following probability:

$P(X\leq 20)$

And we can use the cumulative distribution function given by:

$F(X) =\frac{x-a}{b-a}=\frac{x-10}{35} , 10 \leq X \leq 45$

And we can calculate the probability like this:

$P(X \leq 20) = F(20) = \frac{20-10}{35}=0.286$

Step-by-step explanation:

For this case we define our random variable X="Lindsay's late arrival time, in minutes"

Part a

We know that the distribution for X is given by: $X \sim Unif (a=10, b=45)$

The density function for this distirbution is given by:

$f(x) = \frac{1}{b-a}= \frac{1}{45-10}= \frac{1}{35} , 10 \leq X \leq 45$

So then the height of the uniform density curve would be given by:

$h = \frac{1}{35}= 0.029$

Part b

For this case we want to calculate the following probability:

$P(X\leq 20)$

And we can use the cumulative distribution function given by:

$F(X) =\frac{x-a}{b-a}=\frac{x-10}{35} , 10 \leq X \leq 45$

And we can calculate the probability like this:

$P(X \leq 20) = F(20) = \frac{20-10}{35}=0.286$

7 0

4 years ago