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3 years ago

I need help fast

Mathematics

1 answer:

Anni [7]3 years ago

6 0

Answer:

2.83 units diagonally

2 units horizontally and 2 units vertically

Step-by-step explanation:

x position difference -2-(-4)=2

y position difference 7-5=2

2 units horizontally and 2 units vertically on a graph

2^2+2^2=8

sq rt 8=2.83 units

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4 years ago

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What is the value of log 7 343?–3 -1/3 1/3 3

xeze [42]

Answer: 3

Step-by-step explanation:

Given :

$Log_{7}$ 343

We need to write 343 in index form of 7.

343 is the same as $7^{3}$ , replacing 343 with $7^{3}$ , we have

$Log_{7}$ $7^{3}$

Recall one on the law of Logarithm :

$Log_{a}$ $b^{c}$ can be written as c $Log_{a}$ b

So, $Log_{7}$ $7^{3}$ can be written as ;

3 $Log_{7}$ 7

Also from the laws of Logarithms , $Log_{a}$ a = 1

so , Log_{7}[/tex] 7 = 1

The solution then becomes

3 x 1 = 3

Therefore :

$Log_{7}$ 343 = 3

8 0

3 years ago

An object is launched straight into the air. The projectile motion of the object can be modeled using h(t) = 96t – 16t2, where t

ratelena [41]

Answer with Step-by-step explanation:

We are given that height of projectile after t seconds is given by

$h(t)=96t-16t^2$

a.h(t)=144 ft

$144=96t-16t^2$

$16t^2-96t+144=0$

$t^2-6t+9=0$

$t^2-3t-3t+9=0$

$t(t-3)-3(t-3)=0$

$(t-3)(t-3)=0$

t-3=0

t=3

After 3 s, the height of the project will be 144 feet in the air.

b.h(t)=0

$96t-16t^2=0$

$16t(6-t)=0$

$16t=0\implies t=0$

$6-t=0\implies t=6$

At t=0, the initial position of projectile

At t=6 s , the projectile will hit the ground.

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3 years ago

Can someone solve point-slope lines and show work? I’m confused

Bad White [126]

Answer:

Option (2)

Step-by-step explanation:

Equation of a line passing through a point (x', y') and slope 'm' is given by,

y - y' = m(x - x')

From the question in the picture attached,

Slope of the line 'm' = $-\frac{7}{10}$

Line passes through a point (80, -71).

Equation of the line will be,

y + 71 = $-\frac{7}{10}(x-80)$

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y = $-\frac{7}{10}x-15$

Option (2) will be the answer.

8 0

2 years ago