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Juliette [100K]

3 years ago

6

Please help :) ! I give brainly

2 answers:

Jobisdone [24]3 years ago

4 0

Answer:

a. −5x−1

b. 11x−1

Step-by-step explanation:

vaieri [72.5K]3 years ago

4 0

Answer:

A. -5x-1

B. X-1

Step-by-step explanation: Hope this help :)

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3. There are 6 places set at each table for the big outdoor party.

Ainat [17]

You do 6x4 , The amount of legs on the chairs is irrelevant information, but your answer will be 24 people
24 people

8 0

3 years ago

Could y’all help me please please please

natka813 [3]

Answer:

Your answer is 12

Step-by-step explanation:

Start by multiplying 9 and 1 1/2. Then add 7 1/2. Now divide by 1 3/4

4 0

3 years ago

Which expressions can be used to find m∠ABC? Check all that apply.

valina [46]

Answer:

$\cos^{-1}(\frac{6.3}{9.8})$

$\sin^{-1}(\frac{7.5}{9.8})$

Step-by-step explanation:

Recall the mnemonics SOH-CAH-TOA

The sine ratio is the length of the opposite side expressed over the hypotenuse.

$\sin m\angle ABC=\frac{7.5}{9.8}$

Take the sine inverse of both sides to get;

$m\angle ABC=\sin^{-1}(\frac{7.5}{9.8})$

The cosine ratio is the length of the adjacent side expressed over the hypotenuse.

$\cos m\angle ABC=\frac{6.3}{9.8}$

Take the cosine inverse of both sides to get;

$m\angle ABC=\cos^{-1}(\frac{6.3}{9.8})$

4 0

4 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Which scatterplot represents the data given in the table which shows the number of oranges in bags and the weights of the bags?

lisov135 [29]

Answer:

I would need to see an actual table to answer this question.

4 0

2 years ago

Read 2 more answers