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1 year ago

Help please for these two

Mathematics

1 answer:

evablogger [386]1 year ago

4 0

The measure of angle 2 is 129 degrees

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Tristan is building steps up to a deck that is 156 yards above the ground. The steps must all be the same height. If he wants to

NARA [144]

Answer:

The height of each step is 1957 yards

Step-by-step explanation:

we are given

Tristan is building steps up to a deck that is 15656 yards above the ground

so, total height =15656 yards

Number of steps =8

now, we can use formula

Height of each step = ( total height)/( number of steps)

now, we can plug these values

and we get

Height of each step is

6 0

3 years ago

-3x + y + 8 - 5y + 7x

djyliett [7]

The answer is : 4x-4y+8

5 0

3 years ago

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Which system of equations is represented by the graph? Picture attached

STatiana [176]

None

the slope of the suggested lines is either 1 or -1
and none of the equal to the slope of the red line in the graph wichc is 2

6 0

3 years ago

Read 2 more answers

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

What is the solution to this equation x/5=15<br> a.x=75<br> b.x=20<br> c.x=3<br> d.x=10

Darina [25.2K]

$\dfrac{x}{5}=15\ \ \ \ |\cdot5\\\\\not5\cdot\dfrac{x}{\not5}=5\cdot15\\\\x=75$

Answer: a. x = 75

8 0

3 years ago