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2 years ago

I need some help out with a question!

Mathematics

1 answer:

Alika [10]2 years ago

3 0

Answer:

Here is your answer plz mark as brainliest

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10.) For the natural logarithm function, determine the point on the curve where its associated tangent line passes through the

Lunna [17]

f(x)=lnx

y=f(x)

dy/dx= 1/x

tangent at (x,y) has slope 1/x

eqn of tangent is y = mx + c

since the tangent passes through origin, c=0

substitute y = lnx and m= 1/x to above eqn

lnx = 1

x=e

y=lne=1

4 0

3 years ago

The junior and senior classes at Central High School were asked to choose a destination for a field trip. The results are shown

Ann [662]

Solution:

We are given below frequency table:

Amusement-Park Museum Broadway-Show Total

Juniors 57 21 42 120

Seniors 64 44 58 166

Total 121 65 100 286

We have to find the percentage of surveyed students who chose the amusement park.

From the table, we see there are total 121 students who chose amusement park out of total 286 students.

Therefore, the percentage of surveyed students who chose the amusement park is given below:

$\frac{121}{286}\times 100=42.3\%$

Hence, the option A. 42.30% correct.

3 0

2 years ago

Choose the equation of the vertical line passing through the point (−4, 2).

Degger [83]

Vertical line has no slope. so only look at the x value. x=-4.

8 0

3 years ago

Read 2 more answers

E: f= 3:7<br> f:g= 2:3<br> Work out e:9

Ksenya-84 [330]

Answer:

e f f g

3 7 2 3

6 14 14 21

e;g=6:21

e : g=2 : 7

Step-by-step explanation:

8 0

2 years ago

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

7 0

3 years ago