Use the expression (8x + 2) + (-9x + 7)

Solve the equation 8·4^x+1=1

stepladder [879]

Answer:

See work below

Step-by-step explanation:

$8(4^x)+1=1$

$9(4^x)=1$

$-9$ $-9$

$(4^x)=-8$

$/-8$ $/-8$

And this is where I'm stuck, can someone help me finish?

4 0

3 years ago

Problem Emily convinced her mom to buy a giant box of her favorite cereal. Her mom doesn't think the box will fit on their shelf

Doss [256]

The height of the box should be 40cm

3 0

3 years ago

Find parametric equations for the line through point P(-4,5,2) in the direction of the vector (-6,6,-5)

tigry1 [53]

The vector v (-6, 6, -5) points in the direction of itself, so we start there.

We can capture all points on the line through the origin and the point (-6, 6, -5) by scaling v by an arbitrary real number t.

The line through point P and pointing in the same direction as v is parallel to the other line that passes through the origin. Then the line we want can be obtained by translating the line through the origin by a vector p that points to (-4, 5, 2), so the vector equation for this line is

r (t ) = p + t v

r (t ) = (-4, 5, 2) + t (-6, 6, -5)

r (t ) = (-4 - 6t, 5 + 6t, 2 - 5t )

To get the parametric equations, simply take out the components:

x (t ) = -4 - 6t

y (t ) = 5 + 6t

z (t ) = 2 - 5t

4 0

3 years ago

When running a 100 meter race Bill reaches his maximum speed when he is 45 meters from the starting line and 5 seconds have elap

Mkey [24]

Bill's speed is constant after 5 seconds and 45 meters into the race.

Correct responses:

a. Bill's maximum speed is 8 m/s

b. 24 meters

c. 57 meters

<h3>Methods used to calculate speed and distance traveled</h3>

Given parameters are;

The distance of the race = 100 meter

Distance at which Bill reaches maximum speed = 45 meters

Speed Bill maintains after 5 seconds = The maximum speed

Time at which Bill is 85 meters from the starting line = 10 seconds after start

a. Required;

Bill's maximum speed in meters.

Solution:

Distance Bill runs at maximum speed, d = 85 m - 45 m = 40 m

Time at which Bill runs the 40 m at maximum speed, t = 10 s - 5 s = 5 s

$Speed = \mathbf{\dfrac{Distance}{Time}}$

Therefore;

$Bill's \ maximum \ speed = \dfrac{40 \ m}{5 \ s} = \underline{8 \ m/s}$

b. i. Required:

The distance Bill will run for 3 seconds at the maximum speed;

Solution:

Distance, s = Speed, v × Time, t

The distance traveled at maximum speed in 3 seconds is therefore;

Distance = 8 m/s × 3 s = 24 m

The distance Bill travels in 3 seconds at the maximum speed is 24 meters.

ii) The distance Bill travels at 8 seconds after start, is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 3 seconds = 24 meters

Therefore;

Bill's distance from the starting line 8 seconds after start is 45 meters + 25 meters = 69 meters

c. Bill's distance 6.5 seconds after the start of the race is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 1.5 seconds = 1.5 s × 8 m/s = 12 meters

Bill's distance from the starting line 6.5 seconds after start of the race = 45 meters + 12 meters = 57 meters

Learn more about distance, speed, time, relationship here:

https://brainly.in/question/49075584

6 0

2 years ago

Here is the right photo please help i need this before july 1st thank you

Reil [10]

Answer:

i hope this helps

Step-by-step explanation:

2 $x^{2}$ -16=0

+16 +16

2 $x^{2}$ =16

÷2 ÷2

$x^{2}$ = 8

$\sqrt{x^{2} }$ = $\sqrt{8}$

x= $\sqrt{8}$

-5 $x^{2}$ +9=0

-9 -9

-5 $x^{2}$ =-9

÷-5 ÷-5

$x^{2}$ =1.8

$\sqrt{x^{2} }$ = $\sqrt{1.8}$

x= $\sqrt{1.8}$

6 $x^{2}$ -15=27

+15 +15

6 $x^{2}$ =42

÷6 ÷6

$x^{2}$ =7

$\sqrt{x^{2} }$ = $\sqrt{7}$

x= $\sqrt{7}$

4 0

4 years ago