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3 years ago

13

Kwame rode bicycle for a distance of xkm and walked for another 1/2 hour at a rate of 6km/hour. If Kwame covered a total distanc

e of 10km,find the distance x he covered by the bicycle

1 answer:

ElenaW [278]3 years ago

7 0

1/2 hour at 6 km/ hour = 3 km.

Total distance was 10km, so he rode his bike for 10 - 3 = 7 km

x = 7

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Simplify 7h^2+6h^2+5h^2+6h^2-h^2

iogann1982 [59]

Answer:

23h^2

Step-by-step explanation:

These are all like terms so we add/subtract them from left to right.

7h^2+6h^2+5h^2+6h^2-h^2

= 24h^2 - h^2

= 23h^2 (answer)

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3 years ago

At what point on the graph of y=1/4x^4 is the tangent line parallel to the line 24x -3y=8

Rama09 [41]

Answer:

a) dy/dx = 4/(2y+1)^2.

(b) y = 4/9 x - 14/9

(c) d2y/dx2 = -64/243

Step-by-step explanation:

You have the following equation

(1)

(a) You first derivative implicitly the equation (1) respect to x:

next, you solve the last result for dy/dx:

(2)

(b) The equation for the tangent line is given by:

(3)

with yo = -2 and xo = -1

To find the slope m you use the result of the equation (2), because dy/dx evaluated in (-1,-2) is the slope at such point:

m =

Hence, by replacing in the equation (3) you obtain:

hence, the equation for the tangent line is y = 4/9 x - 14/9

(c) To find d2y/dx2 you derivative the result obtain in the equation (2):

(4)

the second derivative for the point (-1,-2) is obtained by replacing y=-2 and dy/dx=m=4/9 in the equation (4):

hence, d2y/dx2 evaluated in (-1,-2) is -64/243

Step-by-step explanation:

3 0

3 years ago

Cal is measuring temperature changes in four substances over different time periods as part of a school project. Which data show

djyliett [7]

Answer:

C) The third table. K= 5

$C)\frac{10}{2}=\frac{15}{3}=\frac{25}{5}=\frac{40}{8}=k= 5$

Step-by-step explanation:

1) Below there are the missing data:

A proportional relationship through a constant k. It is obtained when we divide:

$k=\frac{y}{x}$

2)In this case, when we divide the temperature (T) by the time (h).

$k=\frac{T}{h}$

3)So, examining the table below we are searching for a ratio k common to all measures (temperatures over hour).

$A) \frac{12}{3}\neq\frac{25}{5}\neq\frac{36}{6}\neq\frac{81}{9}\\B) \frac{22.5}{3}\neq\frac{30}{4}\neq\frac{52.5}{7}\neq\frac{67.5}{8}\\C)\frac{10}{2}=\frac{15}{3}=\frac{25}{5}=\frac{40}{8}=k= 5\\D)\frac{8.5}{2}\neq\frac{22.5}{5}\neq\frac{28.5}{7}\neq\frac{36}{8}$

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3 years ago

With which information can you construct a unique rhombus?

qaws [65]

I believe the correct answer is D.

4 0

3 years ago

Read 2 more answers

Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes

azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

The centre of the ellipse is at the origin and the X axis is the major axis

It passes through the points (-3, 1) and (2, -2)

<h2>To Find:</h2>

The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

$\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

Given that the ellipse passes through the point (-3, 1)

Hence,

$\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1$

Cross multiplying we get,

9b² + a² = 1 ²× a²b²
a²b² = 9b² + a²

Multiply by 4 on both sides,

4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

$\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1$

Cross multiply,

4b² + 4a² = 1 × a²b²
a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

3a²b² = 32b²
3a² = 32
a² = 32/3----(3)

Substituting in 2,

32/3 × b² = 4b² + 4 × 32/3
32/3 b² = 4b² + 128/3
32/3 b² = (12b² + 128)/3
32b² = 12b² + 128
20b² = 128
b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

$\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1$

$\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1$

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>

8 0

3 years ago