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3 years ago

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The height of a triangle is 4 m more than twice the length of the base. the area of the triangle is 35 me^2. find the height of

the triangle

1 answer:

Yuri [45]3 years ago

5 0

The height of a triangle is 4 m more than twice the length of the base. the area of the triangle is 35 me^2. find the height of the triangle

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Vicki ordered an enchilada ($2.75), refried beans ($1.75), rice ($.95), and sparkling water ($1.35). Estimate the cost of her me

PSYCHO15rus [73]

Well the first step is to add all the prices up which adds up to $6.80 and multiply it by 1.15. Since the 1 is for the original price which is 6.8 and the .15 is for the tip. You’re welcome and have a nice day

3 0

3 years ago

A student studying his classmates' cars' ages reports the 5-number summary as {1, 3, 4, 8, 20}. what is the iqr for his data?

timurjin [86]

IQR for the data {1, 3, 4, 8, 20} is 5.

Interquartile Range(IQR) for a data is defined as the difference between third quartile and first quartile.

denoted by IQR=Q₃-Q₁ (Q₃ is third quartile and Q₁ is first quartile)

The five number summary of a data set is represented as

{minimum, first quartile , second quartile , third quartile , maximum}..…(1)

According to the question

{1, 3, 4, 8, 20}

On comparing it with equation …(1)

minimum=1

Q₁ =3 , Q₂=4 , Q₃=8 , maximum =20

IQR=Q₃-Q₁

Substituting the values we get

IQR=8-3=5.

Therefore , Interquartile Range for the data {1, 3, 4, 8, 20} is 5.

Learn more about Inter Quartile Range here brainly.com/question/28037049

#SPJ4

3 0

2 years ago

PLEASE HELP ME ASAP

goblinko [34]

Answer:

20.04 units

Step-by-step explanation:

The diagram shows right triangle ABC. Use the definition of the cosine:

$\cos \angle C=\dfrac{\text{Adjacent leg}}{\text{Hypotenuse}}\\ \\\cos \angle C=\dfrac{BC}{AC}\\ \\\cos 51^{\circ}=\dfrac{BC}{27}\\ \\BC=27\cdot \cos 51^{\circ}\\ \\BC\approx 20.04\ units$

3 0

4 years ago

A regular pentagon is shown. A regular pentagon has side lengths of 9.4 centimeters and a radius of 8 centimeters. What is the l

Angelina_Jolie [31]

Answer:

6.5 cm

Step-by-step explanation:

The computation of the length of the apothem is shown below:

Given that

The side length is 9.4 centimeters

And, the radius is 8 centimeters

Now based on the above information

As per the attached figure

AB = 8 cm

BC = 9.4 ÷ 2 = 4.7 cm

Now apply the pythagoras theorem

AB^2 = BC^2 + AC^2

8^2 = 4.7^2 + AC^2

AC^2 = 41.91

AC = 6.47 cm

= 6.5 cm

7 0

3 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago