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2 years ago

I need help with this question please

Mathematics

1 answer:

Cerrena [4.2K]2 years ago

4 0

Answer:

option b is the correct answer of given statement

Step-by-step explanation:

percent of SSSSD strings

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What is the slope of the following line?

Mila [183]

1/2 is the slope of the following when you simplify

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2 years ago

Your restaurant bill is $28.35. Find the amount of TAX you would pay for your dinner if the tax is 6%. (Round to the nearest cen

hoa [83]

Answer:

The tax is $1.70

Step-by-step explanation:

Divide 28.35 by 100 then multiply by six. This is how you reach 6%

4 0

3 years ago

In the diagram below bc is an altitude of abd.to the nearest whole unit.what is the length of cd

Anestetic [448]

Pythagorean theorem:
                       AB = √30^2 + 16^2 = 34

These are similar triangles.

∠BAC is congruent to ∠BAD
∠ACB is congruent with ∠ABD (both right angles)

Using ratios:
               AC / AB = AB / (AC + CD)
             AC + CD = AB² / AC
                       CD = AB² / AC - AC
                       CD = 34² / 16 - 16
                       CD = 56.25
The answer to this question is 56.

7 0

3 years ago

Read 2 more answers

Using Euler’s Method with a step size of h=0.5, estimate the value for y(3) for the differential equation xy dy/dx= 1+ln(x^2), w

lora16 [44]

Euler's method uses the recurrence relation

$y_{n+1}=y_n+hf(x_n,y_n)$

to approximate the value of the solution $y(x)$ to the ODE $y'=f(x,y)$ .

$xy\dfrac{\mathrm dy}{\mathrm dx}=1+\ln(x^2)\implies y'=\dfrac{1+\ln(x^2)}{xy}=f(x,y)$

With a step size of $h=0.5$ , there will only be two steps necessary to find the approximate value of $y(3)$ based on the initial point $y(2)=5$ . See the attached table below for the computation results.

To demonstrate how the table is generated: Since $y(2)=5$ , you are using $(x_0,y_0)=(2,5)$ .

$y_1=y_0+hf(x_0,y_0)$
$y_1=5+0.5\times\dfrac{1+\ln(2^2)}{2\times5}$
$y_1\approx5.1193$

The next point then uses $x_1=x_0+0.5=2.5$

$y_2=y_1+hf(x_1,y_1)$
$y_1=y_1+0.5\times\dfrac{1+\ln(2.5^2)}{2.5y_1}$
$y_2\approx5.230$

8 0

2 years ago

62394 rounded to the nearest tenth

Minchanka [31]

It would be 62400 that was easy I doubt u needed help with that

8 0

3 years ago

Read 2 more answers