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3 years ago

Every Sunday John goes to church from his home . He move 2 miles east. How much distance is between home and church

Mathematics

1 answer:

GREYUIT [131]3 years ago

8 0

Answer:

2 miles

Step-by-step explanation:

Because I said so

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In parallelogram ABCD, AD = 12 in, m∠C = 46º, m∠DBA = 72º. Find the area of ABCD

dmitriy555 [2]

Answer:

The area of parallelogram ABCD is $78.42 \mathrm{in}^{2}$

Explanation:

Given:

AD = 12 in

$m \angle C=46^{\circ}$

$m \angle D B A=72^{\circ}$

To Find:

The area of parallelogram ABCD=?

Solution:

When we construct the parallelogram with the given data, we get a parallelogram formed by 12 cm as one side and an angle with 46 degrees.

The area of the parallelogram can be calculated by $a b * \sin (a n g l e)$

Substituting the value of a=12 we have

$\text { Area of parallelogram }=12 * \text { bsin } 46$

To find the value of b,

We know that area of a triangle can be expressed as,

$\text { Area of triangle }=(A b / 2) \sin (\text {angle})$

So,

$(12 * B D / 2) * \sin 46=(A B * B D / 2) * \sin 72$

Cancelling BD and 2 on both sides we get,

$12 * \sin 46=A B * \sin 72$

$A B=12 * \frac{\sin 46}{\sin 72}$

Therefore,

$b=\frac{12 \sin 46}{\sin 72}$

Substituting the value of b,

$=12 *\left(\frac{12 \sin 46}{\sin 72}\right) * \sin 46$

=78.42

So the area of the parallelogram is $78.42 \mathrm{in}^{2}$

6 0

4 years ago

Read 2 more answers

Help me with this question please

gladu [14]

He can make 10 more bowls of clay if he wants

7 0

3 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

3 years ago

HELLLLLLLLLPPPPPPP The difference of two numbers is 8 and their quotient is 2

Dmitriy789 [7]

Answer:

x-y =8

x/y=3

=> 3y-y=8

=> 2y = 8

=> y = 4

=> x = 4+8 =12

3 0

3 years ago

Find the midpoint of the segment with the given endpoints. M(–2, 1) and N(–3, 2)

mafiozo [28]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
M(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad 
N(\stackrel{x_2}{-3}~,~\stackrel{y_2}{2})
\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{-3-2}{2}~~,~~\cfrac{2+1}{2} \right)\implies \left(-\frac{5}{2}~,~\frac{3}{2} \right)$

5 0

3 years ago

Read 2 more answers