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Colt1911 [192]
3 years ago
8

20x -10x to the second power + 5x to the third power . In descending order

Mathematics
1 answer:
mario62 [17]3 years ago
7 0
It'd be best if you'd stick to symbols.  My interpretation of what you've typed here is 20x - 20x^2 + 5x^3.  In descending order by powers of x, that gives us

5x^3 - 20x^2 + 20x.
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What is the exact area of triangle with measures a=6 b=7 and cos C= 1/4
Julli [10]

The area of the triangle is A= (1/2)absinγ

(sinγ)Λ2+(cosγ)Λ2=1 => sinγ=√1-(cosγ)∧2= √1- (1/4)∧2=√1-(1/16)=√(15/16)=√15/4

A=(1/2)*6*7*√15/4=21*√15/4=(21/4)√15

Good luck!!!

5 0
3 years ago
4. Use information in the figure below to find M/_ D​
shtirl [24]

Answer:

d

Step-by-step explanation:

Since DF = EF then the triangle is isosceles with base angle being congruent, that is

∠ D = ∠ E , then

∠ D = \frac{180-133}{2} = \frac{47}{2} = 23.5°

8 0
3 years ago
Of 60 students, 24 students have a pet. What is the ratio of students with pets to students without pets?
frosja888 [35]

Answer

Find out the what is the ratio of students with pets to students without pets .

To prove

As given

of 60 students, 24 students have a pet.

Thus Total number of student = 60

students have pets = 24

students without  pets = 60 - 24

                                 = 36

Now  ratio of students with pets to students without pets .

\frac{Student\ with\ pets}{Students\ without\ pets} = \frac{24}{36}

simplify the above

\frac{Student\ with\ pets}{Students\ without\ pets} = \frac{2}{3}

Therefore the  ratio of students with pets to students without pets be 2:3 .

Hence proved

4 0
3 years ago
Read 2 more answers
A butcher is dividing a round of beef into 3 equal sized roasts. If each weighs 7.9 pounds, and there are 1.4 pounds of beef lef
Bas_tet [7]
[7.9x3]+1.4=25.1pound
6 0
4 years ago
A 5.8-ft-tall person walks away from a 9-ft lamppost at a constant rate of 3.4 ft/sec. What is the rate that the tip of the pers
Dovator [93]

Answer:

9.56 ft/sec

Step-by-step explanation:

We are told that a 5.8-ft-tall person walks away from a 9-ft lamppost at a constant rate of 3.4 ft/sec.

I've attached an image showing  triangle that depicts this;

Thus; dx/dt = 3.4 ft/sec

From the attached image and using principle of similar triangles, we can say that;

9/y = 5.8/(y - x)

9(y - x) = 5.8y

9y - 9x = 5.8y

9y - 5.8y = 9x

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y = 9x/3.2

dy/dx = 9/3.2

Now, to find how fast the tip of the shadow is moving away from the lamp post, it is;

dy/dt = dy/dx × dx/dt

dy/dt = (9/3.2) × 3.4

dy/dt = 9.5625 ft/s ≈ 9.56 ft/sec

5 0
3 years ago
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