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ivolga24 [154]
2 years ago
12

Classify each graph as increasing, decreasing, constant

Mathematics
1 answer:
posledela2 years ago
8 0

Answer:

3. increasing

4. decreasing

Step-by-step explanation:

well for graph 3 the values are increasing

and for graph 4 the values are decreasing

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Evaluate the expression into simplest form
sattari [20]
Work  out  the top subtraction first:

3/9 -  8/12

= 1/3 - 2/3

= 1/3


and the bottom part:-

3/8 * 2 = 6/8 = 3/4

so now we divide -1/3 by 3/4

= -1/3 * 4/3  = -4/9

5 0
3 years ago
In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?
7nadin3 [17]

Correct answer is: distance from D to AB is 6cm

Solution:-

Let us assume E is the altitude drawn from D to AB.

Given that m∠ACB=120° and ABC is isosceles which means

m∠ABC=m∠BAC = \frac{180-120}{2}=30

And AC= BC

Let AC=BC=x

Then from ΔACD , cos(∠ACD) = \frac{DC}{AC} =\frac{4}{x}

Since DCB is a straight line m∠ACD+m∠ACB =180

                                              m∠ACD = 180-m∠ACB = 60

Hence cos(60)=\frac{4}{x}

          x=\frac{4}{cos60}= 8

Now let us consider ΔBDE, sin(∠DBE) = \frac{DE}{DB} =\frac{DE}{DA+AB} = \frac{DE}{4+8}

DE = 12sin(30) = 6cm

7 0
4 years ago
Please help ill give brainliest to right answer
ch4aika [34]

Answer: D y:2/5x

It correct

7 0
3 years ago
2+2= HURRY I THINK ITS GOLDFISH BUT I DONT KNOW
Bingel [31]

Answer:

4

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Find the sum of the geometric series.<br><br> a <br> b <br> c <br> d
Musya8 [376]

Answer:

B. 4\sqrt{3}-6

Step-by-step explanation:

We have,

The first term of the series, a=\sqrt{3}.

The common difference is given by, r=\frac{\frac{-3}{2}}{\sqrt{3}} i.e. r=\frac{-\sqrt{3}}{2}.

Since, the given series is an infinite series, then,

Sum of an infinite series = \frac{a}{1-r}

i.e. Sum the series = \frac{\sqrt{3}}{1+\frac{\sqrt{3}}{2}}

i.e. Sum the series = \frac{2\sqrt{3}}{2+\sqrt{3}}

i.e. Sum the series = \frac{2\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}

i.e. Sum the series = \frac{2\sqrt{3}\times (2-\sqrt{3})}{4-3}

i.e. Sum the series = 4\sqrt{3}-6

Thus, the sum of the series is 4\sqrt{3}-6.

8 0
3 years ago
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