PLEASE HELP ME!! quick geometry!!

3. In class today, 1/4 of the students wore shorts and

Mkey [24]

Answer:

A. $\frac{1}{4}$ = $\frac{3}{12}$

Step-by-step explanation:

$\frac{1}{4}$ of the students wore shorts.

$\frac{3}{12}$ of the students wore jeans.

Converting the fractions of the students who wore shorts and jeans respectively;

$\frac{1}{4}$ = 0.25

$\frac{3}{12}$ = 0.25

This means that the ratio of students who wore shorts is the same as that of students who wore jeans.

i.e $\frac{1}{4}$ = $\frac{3}{12}$

4 0

2 years ago

Read 2 more answers

Damon has five dimes and some nickels in his pocket, worth of a total of $1.20. To find the number of nickels Damon has a, stude

Agata [3.3K]

The answer to the equation should be in cents whereas it is in dollars which would be incorrect

6 0

2 years ago

Experimentally verify sum of three angles of an triangle is 180⁰.

svetlana [45]

Draw line a through points A and B. Draw line b through point C and parallel to line a.

Since lines a and b are parallel, <)BAC = <)B'CA and <)ABC = <)BCA'.

It is obvious that <)B'CA + <)ACB + <)BCA' = 180 degrees.

Thus <)ABC + <)BCA + <)CAB = 180 degrees.

Lemma

If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.

Proof

Assume to the contrary that AB and DC are not parallel.

Draw a line trough A and B and draw a line trough D and C.

These lines are not parallel so they cross at one point. Call this point E.

Notice that <)AEC is greater than 0.

Since <)CAB = <)DCA, <)CAE + <)ACE = 180 degrees.

Hence <)AEC + <)CAE + <)ACE is greater than 180 degrees.

Contradiction. This completes the proof.

Definition

Two Triangles ABC and A'B'C' are congruent if and only if

|AB| = |A'B'|, |AC| = |A'C'|, |BC| = |B'C'| and,

<)ABC = <)A'B'C', <)BCA = <)B'C'A', <)CAB = <)C'A'B'.

4 0

2 years ago

Sam borrows $5700 at 4.5% simple interest for 3 years. Find the interest

kherson [118]

Answer:

The interest is

Step-by-step explanation:

Simple interest is given by

$I = \frac{P \times R \times T}{100}$

where

P is the principal

R is the rate

T is the time given

From the question

The principal / P = $ 5700

The rate / R = 4.5%

The time given / T = 3 years

So the interest is

$I = \frac{5700 \times 4.5 \times 3}{100}$

$I = \frac{76950}{100}$

We have the final answer as

I = $ 769.50

Hope this helps you

3 0

2 years ago

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$

We use the initial condition $y(0)=-12$ to find the value of C.

$-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}$

Therefore,

$y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}$

4 0

2 years ago