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2 years ago

SOMEONE PLEASE HELP‼️‼️‼️ ILL MARK BRAINLYEST PLEASE ‼️‼️

Mathematics

2 answers:

vladimir2022 [97]2 years ago

7 0

Maybe 1/3 because a triangle has 3 sides?

ki77a [65]2 years ago

5 0

Answer: 1/2

Step-by-step explanation:

No matter which midsegment you created, it will be one-half the length of the triangle's base (the side you did not use), and the midsegment and base will be parallel lines!

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Please help due soon!!!

butalik [34]

Hey there!

Question #1.
2^6 + (2^3)^3

= 64 + (8)^3

= 64 + 8^3

= 64 + 512

= 576

Therefore, the answer should be:

Option A. 576

Question #2.
11^-4 * 11^8

= 1/14,641 * 214,358,881

= 14,641

≈ 11^4

Therefore, the answer should be:

Option C. 11^4

Good luck on your assignment & enjoy your day!

~Amphitrite1040:)

3 0

2 years ago

Read 2 more answers

Factor then solve. 3x2+14x+8=0

tangare [24]

Answer:

x = - 4, x = - $\frac{2}{3}$

Step-by-step explanation:

Given

3x² + 14x + 8 = 0

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term

product = 3 + 8 = 24 and sum = + 14

The factors are + 12 and + 2

Use these factors to split the x- term

3x² + 12x + 2x + 8 = 0 ( factor the first/second and third/fourth terms )

3x(x + 4) + 2(x + 4) = 0 ← factor out (x + 4) from each term

(x + 4)(3x + 2) = 0

Equate each factor to zero and solve for x

x + 4 = 0 ⇒ x = - 4

3x + 2 = 0 ⇒ 3x = - 2 ⇒ x = - $\frac{2}{3}$

4 0

3 years ago

I don't understand this math. 2y^2-5+1+2y^2-6y^2-3-4y

julsineya [31]

Answer:

-2y^2 - 4y - 7

Step-by-step explanation:

If what you need to do is to combine like terms, then do this:

2y^2 - 5 + 1 + 2y^2 - 6y^2 - 3 - 4y =

= 2y^2 + 2y^2 - 6y^2 - 4y - 5 + 1 - 3

= -2y^2 - 4y - 7

6 0

3 years ago

Read 2 more answers

15 points

kirza4 [7]

$_____________________________________$

<h3>ANSWER:</h3>

$\tt\large{y=\frac{9}{2}}$

$\tt\large{x=-5+y}$

$_____________________________________$

<h3>꧁____YunaKoharuChan____꧂</h3>

4 0

3 years ago

Writing g for the acceleration due to gravity, the period,T, of a pendulum of length l is given by

ivanzaharov [21]

Step-by-step explanation:

$T=2\pi\sqrt{\frac{l}{g}}$

$T+\Delta T=2\pi\sqrt{\frac{(l+\Delta l)}{g}} =2\pi\sqrt{\frac{l}{g}}\sqrt{1+\frac{\Delta l}{l}}=2\pi\sqrt{\frac{l}{g}}(1+\frac{1}{2}\frac{\Delta l}{l}+0(\frac{\Delta l}{l}))=T(1+\frac{1}{2}\frac{\Delta l}{l}+0(\frac{\Delta l}{l}))$

Here, the Taylor approximation for a square root was applied, and O(x) stands for all negligible terms of Taylor's sum with respect to variable x.

So, $\Delta T=T\frac{1}{2}\frac{\Delta l}{l}$

b. For an increase of 2%, that is:

$\frac{\Delta l}{l}=0.02$

$\frac{\Delta T}{T}=\frac{1}{2}0.02=0.01=1\%$

7 0

3 years ago

Read 2 more answers