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2 years ago

(8.4C) Ken has a weekly goal for the number of

Mathematics

1 answer:

77julia77 [94]2 years ago

3 0

Answer:

the slope is -300

Step-by-step explanation:

y=mx+b

m is slope

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6) R E 126 10x + 8 13 + 5x D A) 1 C) 5 B) 7 D) 14

Gemiola [76]

Step-by-step explanation:

step 1. what is the question? maybe the question is what is the value of x in the diagram. okay.

step 2. interior angles of a triangle add up to 180°

step 3. the 3 angles inside the triangle are 10x + 8, 13 + 5x, and (180 - 126)

step 4. (10x + 8) + (13 + 5x) + 54 = 180

step 5. 15x + 21 = 126

step 6. 15x = 105

step 7. x = 7.

3 0

2 years ago

Omplete the statement. Round to the nearest hundredth if necessary. 75.2 in. ≈≈ cm

salantis [7]

The total is 191.008 centimeters.

4 0

2 years ago

Solve for x 4r – 9 – 6x = 21

ehidna [41]

Answer:

X= 30-4r/-6

Step-by-step explanation:

1. Move the 9 to the other side. (21+9=30)

2. subtract the 4r and move it to the other side (-6x=21-4r)

3. divide both sides by -6

5 0

2 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

Write an equation of a line that has a slope of –5 and a y-intercept of 2.

Bingel [31]

Answer:

y-2 = -5x

5x+y = 2 is the required equation of a line.

4 0

2 years ago