All categories

2 years ago

Please answer & explain how the answer was found

Mathematics

1 answer:

pshichka [43]2 years ago

3 0

Answer: GEF = 30 DEF = 40

Step-by-step explanation: By angle summation we know the entire angle

GED = 70, so the smaller angles that make up GED (GEF and DEF) will have to sum up to equal GED. Using this we get:

2x + 10 + 5x - 10 = 70

From here we can combine like terms:

7x = 70

and solve for x

x = 10

Then we can plug the value of x back into the equations given to solve for GEF and DEF

GEF:

2x+10 == 2(10)+10 = 30

DEF:

5x-10 == 5(10)-10 = 40

You might be interested in

3. Mr. James told Jill he would charge her $15 per hour, plus parts, to fix her bicycle.

ad-work [718]

Answer:

512

Solution:

so first you multiply 15 times 32 and get 480 than you add 32 and get 512.

7 0

3 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago

Plz help. I'll give brainliest to whoever helps me answer this

Rzqust [24]

Answer: $34.50

Step-by-step explanation

If you multiply .75=3/4 x 10 it will will be 7.50 so 10x 3.45= 34.50 or 7.50 x 3.45= 25.875

25.875/.75 = $34.50

3 0

3 years ago

The volume of ice-cream in the cone is half the volume of the cone. The cone has a 3 cm radius and

zhannawk [14.2K]

Answer:

h = 5 cm

Step-by-step explanation:

Given that,

The volume of ice-cream in the cone is half the volume of the cone.

Volume of cone is given by :

$V_c=\dfrac{1}{3}\pi r^2h$

r is radius of cone, r = 3 cm

h is height of cone, h = 6 cm

So,

$V_c=\dfrac{1}{3}\pi (3)^2\times 6\\\\V_c=18\pi\ cm^3$

Let $V_i$ is the volume of icecream in the cone. So,

$V_i=\dfrac{18\pi}{2}=9\pi\ cm^3$

Let H be the depth of the icecream.

Two triangles formed by the cone and the icecream will be similiar. SO,

$\dfrac{H}{6}=\dfrac{r}{3}\\\\r=\dfrac{H}{2}$

So, volume of icecream in the cone is :

$V_c=\dfrac{1}{3}\pi (\dfrac{h}{2})^2(\dfrac{h}{3})\\\\9\pi=\dfrac{h^3}{12}\pi\\\\h^3=108\\\\h=4.76\ cm$

h = 5 cm

So, the depth of the ice-cream is 5 cm.

8 0

3 years ago

What is 9/10 + 7/15

Yakvenalex [24]

$\huge\text{Hey there!}$

$\mathsf{\dfrac{9}{10}+\dfrac{7}{15}}$

$\large\textsf{FIRST: FIND the LCD (Lowest Common Denominator) then solve}\\\large\textsf{from there!}$

$\large\textsf{If you have calculated it correctly, you should have came up with \underline{\bf 30}}\\\large\textsf{as your LCD (Lowest Common Denominator).}$

$\mathsf{= \dfrac{9\times3}{10\times3}+ \dfrac{7\times2}{15\times2}}$

$\mathsf{9\times3=\bf 27}\\\mathsf{10\times3=\bf 30}\\\\\mathsf{7\times2=\bf 14}\\\mathsf{15\times2=\bf 30}$

$\mathsf{= \dfrac{27}{30}+\dfrac{14}{30}}$

$\mathsf{= \dfrac{27+14}{30}}$

$\mathsf{27+ 14=\bf 41}\\\\\mathsf{30+0=\bf 30}$

$\mathsf{= \dfrac{41}{30}}\large\textsf{ which you could convert to }\mathsf{1 \dfrac{11}{30}}$

$\boxed{\boxed{\large\textsf{ANSWER: }\bf \dfrac{41}{30} \large\textsf{ or }\mathsf{\bf 1 \dfrac{11}{30}\large\textsf{ because they both equal the same thing}}}}}\huge\checkmark$

$\large\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

6 0

2 years ago

Read 2 more answers