All categories

3 years ago

Basing your answer on the appearance of the figures below, identify whether the mathematical expression is true or false.

Mathematics

2 answers:

weeeeeb [17]3 years ago

5 0

Answer : The mathematical expression is false.

Step-by-step explanation :

As we are given 4 triangles in which ΔROB & ΔPTA and ΔDEF & ΔYXW are appears congruent.

First we have to show that ΔROB and ΔPTA appear congruent.

Side RO appears equal to Side PT

Side OB appears equal to Side TA

Side RB appears equal to Side PA

∴ ΔROB ≅ ΔPTA (by SSS)

Now we have to show that ΔDEF and ΔYXW appear congruent.

Side DE appears equal to Side YX

Side EF appears equal to Side XW

Side DF appears equal to Side YW

∴ ΔDEF ≅ ΔYXW (by SSS)

According to given expression, ΔROB and ΔDEF have different shapes and sizes.

So, ΔROB not appear congruent to ΔDEF

Therefore, the mathematical expression is false.

KatRina [158]3 years ago

4 0

Answer:

The mathematical expression is false

Step-by-step explanation:

* Lets use the figure to answer the question

- There are four triangles in the figure

- Δ ROB and Δ PTA appear congruent because:

# The side RO appears equal the side PT

∴ RO ≅ PT

# The side OB appears equal the side TA

∴ OB ≅ TA

# The side RB appears equal the side PA

∴ RB ≅ PA ⇒ SSS

∴ Δ ROB ≅ Δ PTA

- Δ DEF and Δ YXW appear congruent because:

# The side DE appears equal the side YX

∴ DE ≅ YX

# The side EF appears equal the side XW

∴ EF ≅ XW

# The side DF appears equal the side YW

∴ DF ≅ YW

∴ Δ DEF ≅ Δ YXW ⇒ SSS

- Δ ROB and Δ DEF have different shapes and sizes

∵ Δ ROB not appear congruent to Δ DEF

∴ Δ ROB ≠ Δ DEF

∴ The mathematical expression is false

You might be interested in

Five children have heights of 138cm, 135cm, 140cm, 139cm, and 141cm.

Bess [88]

Answer:

6 cm

Step-by-step explanation:

To find the range take the largest number and subtract the smallest number

The largest number is 141 and the smallest number is 135

141 - 135 = 6 cm

8 0

3 years ago

Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea

Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) $p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

d) $0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

$P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}$

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

$p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}$

And if we simplify we got this:

$p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we simplify we got:

$p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

$p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And if we set equal to 0 we got:

$0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

7 0

3 years ago

Is 7.8 cm, 1.1 cm, 9.1 cm a triangle

Anna35 [415]

Answer: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Since, 11 + 21 > 16, 11 + 16 > 21, and 16 + 21 > 11, you can form a triangle with side lengths 11 mm, 21 mm, and 16 mm.

Step-by-step explanation:

5 0

2 years ago

Is 0.27 greater than 0.3 ?

Zinaida [17]

Answer:

Step-by-step explanation:

here's a little trick, add a zero to make both numbers the same.

it's doesn't change the value.

so .27 and .3 become .27 and .30

and .30 is greater

7 0

3 years ago

In an experiment, the population of bacteria is increasing

Nezavi [6.7K]

25 million

number of bacteria doubles every minute

6 0

2 years ago