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3 years ago

Y = −4x + 3 Graphing equations in slope-intercept form

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

4 0

Y=-4x+3 is already in slope intercept form. Graphed it looks like this

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100m In 9 3/5 seconds

Nostrana [21]

Answer:

Speed = 10.42 m/secs

Step-by-step explanation:

We have been given with:-

Distance = 100m

Time taken = 9 $\frac{3}{5}$ secs

1) First, we need to convert the mixed number 9 $\frac{3}{5}$ to an improper fraction.

Method to convert a mixed number to an improper fraction

 $a\frac{b}{c} = \frac{c*a+b}{c}$ 

Applying the above formula, we get

9 $\frac{3}{5}$ = $\frac{5*9+3}{5}$

= $\frac{45+3}{5}$

= $\frac{48}{5}$

2) Now we need to convert $\frac{48}{5}$ into a decimal by dividing 48 by 5.

We get time taken as 9.6 secs

3) We need to find the speed using the below formula:-

Speed = $\frac{Distance}{Time taken}$

Plugging the values of Distance and Time taken into the formula, we get

Speed = $\frac{100}{9.6}$

= 10.416 m/secs

4) Upon rounding it to the nearest hundredth, we get

Speed = 10.42 m/secs

4 0

3 years ago

Please answer this correctly as soon as possible I want genius or expert people to answer this correctly

Ede4ka [16]

6/54 times because there are 9 marbles and the orange marble is 1 out of those 9 marbles. If you multiply 6*9= 54 then you notice there is a 6/54 chance of picking any 1 marble out of the 54 times that you decide to pick marbles.

4 0

3 years ago

Please help me with these, oh sweet jesus

Lelechka [254]

Answer:

77. $\cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} x$ Proved

78. $\sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ]$ Proved

79. $\cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x$ Proved.

80. $\sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x$ Proved.

Step-by-step explanation:

77. Left hand side

= $\cot^{6} x$

= $\cot^{4} x \times \cot^{2} x$

= $\cot^{4}x [\csc^{2}x - 1]$

{Since we know, $\csc^{2} x - \cot^{2}x = 1$ }

= $\cot^{4} x \csc^{2}x - \cot^{4} x$

= Right hand side (Proved)

78. Left hand side

= $\sec^{4}x \tan^{2} x$

= $\sec^{2} x [1 + \tan^{2}x] \tan^{2} x$

{Since $\sec^{2}x - \tan^{2}x = 1$ }

= $\sec^{2}x [\tan^{2}x + \tan^{4}x ]$

= Right hand side (Proved)

79. Left hand side

= $\cos^{3} x\sin^{2} x$

= $\cos x[1 - \sin^{2} x] \sin^{2} x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $[\sin^{2}x - \sin^{4}x] \cos x$

= Right hand side

80. Left hand side

= $\sin^{4}x - \cos^{4}x$

= $[\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $1 - 2\cos^{2} x[1 - \cos^{2}x ]$

= $1 - 2\cos^{2}x + 2 \cos^{4} x$

= Right hand side. (Proved)

7 0

3 years ago

Wats 3/500 as a percent

Mrac [35]

Your answer is 0.6%.

5 0

3 years ago

Read 2 more answers

On section AB, whose length is 192 cm, take the point C so that AC: CB = 1: 3. The point D is taken on the AC section so that CD

jasenka [17]

Answer:

112 cm

Step-by-step explanation:

Given:

AB = 192 cm
AC : CB = 1 : 3
CD = BC/12
The distance between midpoints of AD and CB = x

Find the length of AC and CB:

AC + CB = AB
AC + 3AC = 192
4AC = 192
AC = 192/4
AC = 48 cm

Find CB:

CB = 3AC = 3*48 = 144 cm

Find the length of CD:

CD = BC/12 = 144/12 = 12 cm

Find the length of AD:

AD = AC - CD = 48 - 12 = 36 cm

Find the midpoint of AD:

m(AD) = 36/2 = 18 cm

Find the midpoint of CB:

m(CB) = AC + 1/2CB = 48 + 144/2 = 48 + 82 = 130 cm

Find the distance between the midpoints:

x = 130 - 18 = 112 cm

4 0

2 years ago