All categories

3 years ago

.Answer pls. question in photo

Mathematics

2 answers:

JulsSmile [24]3 years ago

8 0

Answer:56

Step-by-step explanation:

slamgirl [31]3 years ago

6 0

Answer:

+7=36

Step-by-step explanation:

You might be interested in

Can someone help me pls, 50 points

Reptile [31]

Answer:

x = -0.25

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Solve for x given that <br><img src="https://tex.z-dn.net/?f=%20%7B64%7D%5E%7B2x%20-%201%7D%20%3D%20%20%7B32%7D%5E%7Bx%20%2B%201

marissa [1.9K]

Answer:

x = $\frac{11}{7}$

Step-by-step explanation:

note that 64 = $2^{6}$ and 32 = $2^{5}$ , thus

$(2^{6}) ^{2x-1}$ = $(2^{5}) ^{x+1}$ , that is

$2^{6(2x-1)}$ = $2^{5(x+1)}$

$2^{12x-6}$ = $2^{5x+5}$

Since the bases on both sides are equal, then equate the exponents

12x - 6 = 5x + 5 ( subtract 5x from both sides )

7x - 6 = 5 ( add 6 to both sides )

7x = 11 ( divide both sides by 7 )

x = $\frac{11}{7}$

8 0

4 years ago

(Sorry This isn't math, couldn't find science)

vladimir2022 [97]

Answer:

The answer is A

Step-by-step explanation:

I remember looking this up because my gamertag on xbox is Auroras Lights

What happens is our ozone layer is thinning out each year and solar wind particles come in our atmosphere where they mix with the ions. This creates a colorful light show in the sky that we see. It happens at the poles because of a magnetic pull.

7 0

3 years ago

In Which Quadrant is this true

34kurt

Given:

$\sin \theta$

$\tan \theta$

To find:

The quadrant in which $\theta$ lie.

Solution:

Quadrant concept:

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II, only $\sin\theta$ and $\csc\theta$ are positive.

In Quadrant III, only $\tan\theta$ and $\cot\theta$ are positive.

In Quadrant IV, only $\cos\theta$ and $\sec\theta$ are positive.

We have,

$\sin \theta$

$\tan \theta$

Here, $\sin\theta$ is negative and $\tan\theta$ is also negative. It is possible, if $\theta$ lies in the Quadrant IV.

Therefore, the correct option is D.

5 0

3 years ago

In a triathlon, Jenny swam for 1 hour, biked for 1.75 hours, and ran for 1 hour. Her average biking speed was 2 times her averag

expeople1 [14]

Answer:

28r + 8r + r = 55

Step-by-step explanation:

Some values you know (given or calculated) and some things you do not know.

RATE TIME DISTANCE

BIKE 2*8r 1.75

RUN 8r

SWIM r 1

Total 55

Give a variable to unknown running speed and fill in everything.

RATE TIME DISTANCE

BIKE 16r 1.75 28r

RUN 8r t 8rt

SWIM r 1 r

Total 55

Notice some trouble? WHY? I missed seeing the "ran for one hour". the variable t is KNOWN! This will be corrected here:

RATE TIME DISTANCE

BIKE 16r 1.75 28r

RUN 8r 1 8r

SWIM r 1 r

Total 55

Calculations used to fill out the graph:

Then her running speed was 8r,

and her biking speed was 2*(8r) = 16r.

Swimming for 1 hour, Jenny covered the distance of 1*r = r kilometers.

Biking for 1.75 hour, Jenny covered the distance of 1.75*16r = 28r kilometers.

Running for 1 hour, Jenny covered the distance of 1*2r = 8r kilometers.

Everything is filled and looks good. There is a sum of distances in just the one single variable, r.

Solve for r, and this is the swimming speed.

28r + 8r + r = 55

4 0

3 years ago