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2 years ago

What is the factored form of 27 - 530?

Mathematics

1 answer:

fenix001 [56]2 years ago

8 0

Answer:

1,2,3

Step-by-step explanation:

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29. Find the height of the trapezoid. Use this height to find the area. (A =

harkovskaia [24]

Answer:

<em>The area of the trapezium is 168</em>

Step-by-step explanation:

<u>Area of a Trapezoid</u>

Given a trapezoid of parallel bases b1 and b2, and height h, the area is calculated with the formula:

$\displaystyle A=\frac{b_1+b_2}{2}h$

The trapezoid in the figure has b1=15 and b2=27. We need to find the height. If we focus on triangle BCD, we can calculate the height as the distance EC by using the Pythagora's Theorem:

$10^2=EC^2+BC^2$

The side BC can be found as half the difference of the bases:

$BC=\frac{27-15}{2}=6$

Solving for EC:

$EC^2=10^2-6^2=100-36=64$

$EC=\sqrt{64}=8$

Now we have the height, calculate the area:

$\displaystyle A=\frac{15+27}{2}*8$

$\displaystyle A=\frac{42}{2}*8$

A = 168

The area of the trapezium is 168

5 0

2 years ago

The square pqrs is shown.

ira [324]

Answer:

1:1

Step-by-step explanation:

The ratio would be 1:1 since it's a square, the sides will always be the same. In a square ,all sides are the same length. If they are not, then it wouldn't be a square.

5 0

2 years ago

In ANOP, the measure of Angle P=90°, PN = 89 feet, and OP = 74 feet. Find the measure

padilas [110]

Answer:

O = 50.3 degrees

Step-by-step explanation:

tan O = 89/74

tan O = 1.2027

O = 50.3 degrees

8 0

3 years ago

Read 2 more answers

Which of the following is equal to the absolute value of 7 A.-3and-4 on a number line. B.7and0 on a number line. C.1and7 on a nu

lozanna [386]

Absolute Value can’t be negative so -3 minus -4 is -7 but since absolute value isn’t able to be negative it turns into 7, the answer is A

5 0

2 years ago

Read 2 more answers

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

6 0

2 years ago