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3 years ago

If theperimeter of the triangle ABC is 23, find the length of each side.

Mathematics

1 answer:

aalyn [17]3 years ago

6 0

You can't.
There isn't enough evidence/proof to prove the length of each side.

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Consider two functions f and g on [1, 8] such that integral^8_1 f(x) dx = 9, integral^8_1 g(x) dx = 5, integral^8_5 f(x) dx = 4,

Gelneren [198K]

I'll abbreviate the definite integral with the notation,

$I(f(x),a,b)=\int_a^bf(x)\,\mathrm dx$

We're given

$I(f,1,8)=9$
$I(g,1,8)=5$
$I(f,5,8)=4$
$I(g,1,5)=3$

Recall that the definite integral is additive on the interval $[a,b]$ , meaning for some $c\in[a,b]$ we have

$I(f,a,b)=I(f,a,c)+I(f,c,b)$

The definite integral is also linear in the sense that

$I(kf+\ell g,a,b)=kIf(a,b)+\ell I(g,a,b)$

for some constant scalars $k,\ell$ .

Also, if $a\ge b$ , then

$I(f,a,b)=-I(f,b,a)$

a. $I(2f,1,5)=2I(f,1,5)=2(I(f,1,8)-I(f,5,8))=2(9-4)=\boxed{10}$

b. $I(f-g,1,8)=I(f,1,8)-I(g,1,8)=9-5=\boxed{4}$

c. $I(f-g,1,5)=I(f,1,5)-I(g,1,5)=\dfrac{I(2f,1,5)}2-I(g,1,5)=10-3=\boxed{7}$

d. $I(g-f,5,8)=I(g,5,8)-I(f,5,8)=(I(g,1,8)-I(g,1,5))-I(f,5,8)=(5-3)-4=\boxed{-2}$

e. $I(7g,5,8)=7I(g,5,8)=7(5-3)=\boxed{14}$

f. $I(3f,5,1)=3I(f,5,1)=-3I(f,1,5)=-\dfrac32I(2f,1,5)=-\dfrac32(10)=\boxed{-15}$

4 0

2 years ago

SOMEONE HELP PLEASE!!!

Oliga [24]

Answer:

x = 2 1/3

y = 14 / (2 1/3) = 14 / (7/3) = 14*3/7 = 42/7 = 6

y = 6

x = 4 1/5

y = 14 / (4 1/5) = 14 / (21/5) = 14*5/21 = 70/21 = 3 7/21 = 3 1/3

y = 3 1/3

x = 7/6

y = 14 / (7/6) = 14*6 / 7 = 12

y = 12

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Which point is a reflection of Z(5 1/2,3) across the y axis

RSB [31]

Answer: C?

Step-by-step explanation:

Because reflecting means like think of a mirror it’s reflecting you so the opposite side would be the bottom right quadrant please tell me if I’m right?

5 0

2 years ago

WILL VOTE BRAINLIEST TO THE FIRST CORRECT ANSWER

KonstantinChe [14]

Answer:

Step-by-step explanation:

The D answer choice is the correct functions needed to create the pictured graph, I have pictured these functions in a graphing calculator below

7 0

1 year ago

We looking for distance

drek231 [11]

Answer:

240 m

Step-by-step explanation:

Figure is attached.

The distance the particle travels is the area under the v-t graph over the time interval from 0 min to 4 min. The reason you can't just multiply velocity by time is that the velocity is changing at a constant rate over the given time interval (the line is going upward--the particls is speeding up).

What's the area under the graph?

The red outline forms a trapezoid. The area of a trapezoid is $A=\frac{1}{2}h(b_1+b_2)$ where the b's are the lengths of the two parallel base and h is the height (the distance between the bases).

Reading the graph carefully (notice the scales!)...

$A=\frac{1}{2}(4)(30+90)=(2)(120)=240 \text{m}$

8 0

2 years ago