Kendra bought a magazine for $3 and four paperback books for $5 each the expression 3 + 4 x 5 represents the total cost in dollars of her purchases what are the terms in this expression

It takes approximately 60/7 seconds for the population to grow by one person.

To find out how many seconds it takes for the population to grow by one person, we need to determine the difference between the rate of births and the rate of deaths. The rate of population growth per second can be calculated by subtracting the rate of deaths per second from the rate of births per second. In this case, a person is born every 5 seconds and a person dies every 12 seconds. So, the rate of population growth per second is 1/5 - 1/12 = 12/60 - 5/60 = 7/60.

Therefore, it takes approximately 60/7 seconds for the population to grow by one person.

Answer: (8,-6)

Step-by-step explanation:

the coordinates are  (3,-2)

(x+5, y-4)

(3+5,-2-4)

(8,-6)

If triangle ABC is a right triangle with ∠C = 90°, then angles A and B must be complementary.

If triangle ABC is a right triangle with ∠C = 90°, then angles A and B must be complementary. This means that the sum of angles A and B will be equal to 90°. Therefore, the correct answer is option b.

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The true statement about angles A and B is that a. Angles A and B are complementary.

Since triangle ABC is a right triangle with m∠C = 90°, we know that m∠A + m∠B = 90°.

This means that the sum of the measures of angles A and B is 90 degrees.

So, angles A and B are complementary angles.

In other words, if two angles are complementary, then the sum of their measures is equal to 90 degrees.

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Question

if aabc is a triangle with mlc = 90, which of the following must be true about angles a and b?

Angles A and B are complementary.

Angle B is greater than angle A.

Angles A and B are supplementary

Angle A is greater than angle B.

Answer:

140

Step-by-step explanation:

that's the answer on ttm

Answer:

C. commutative property

Step-by-step explanation:

if you played the game 1000​ times, [tex]\$ 263.20[/tex] are expected to​ lose.

[tex]Probability=\dfrac{Favourable\; outcomes}{Total\; number\; of\; outcomes}[/tex]

Probability of winning[tex]=\dfrac{1}{38}[/tex]

if out of total [tex]38[/tex], [tex]1[/tex] game has been won, so rest have been lost,so

Probability of losing [tex]=\dfrac{37}{38}[/tex]

Hence Expected of the game to the player  [tex]=175 \times \dfrac{1}{38}-5 \times \dfrac{37}{38}[/tex]

                                                                        [tex]=\dfrac{175}{38}- \dfrac{185}{38}\\\\=\$ -0.2632[/tex]

A)The expected value is [tex]\$ -0..2632[/tex] (is losing)

B)if you played the game [tex]1000[/tex] times,

so it will be [tex]1000 \times \$ -0.2632=\$263.20[/tex]

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Applying the multiplication principle in probability, there are eight different combinations possible when choosing between bus or train, full fare or concession, morning or afternoon for your travel.

Your question is asking how many different combinations of travel mode, fare type, and time of day are possible under given conditions. For this situation, we can apply simple probability rules with concept of multiplication principle, which states that if there are n ways to do something, and m ways to do another, then there are n * m ways to do both.

In this case, you have two choices for each category: mode of transportation (bus or train), fare type (full fare or concession), and time of day (morning or afternoon). The combination is simply 2 (mode of transport) * 2 (fare type) * 2 (time of day) = 8. So, there are eight possible outcomes.

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For the first part, part a, we are given that:

0.01b^2+ ∗ +100c^2

In quadratic equations we know that the middle term (the one that is missing) will be twice the square root of the first and the last terms.

Now, we know that the square root of the first term is: [tex] \sqrt{0.01b^2} =0.1b [/tex] Let us represent this by [tex] x [/tex]. Therefore, [tex] x=0.1b [/tex]

and the square root of the last term is: [tex] \sqrt{100c^2} =10c [/tex]

Let this be [tex] y [/tex]. Therefore, [tex] y=10c [/tex]

Therefore, the middle term, * is: [tex] 2xy=2\times0.1b\times 10c=2bc [/tex]

Likewise in the part b, we will have the middle term to be:

[tex] 2\times \sqrt{25a^2} \times \sqrt{\frac{1}{4}b^2} =2\times 5a\times \frac{1}{2}b=5ab [/tex]

because the first and the last terms are 25a^2 and 1/4 b^2 and we know that the middle term will be twice the square root of the first and the last terms.

Answer: ITS NOT 1/6, ITS 1/12

Step-by-step explanation:

I took the test :)

The content of care package would be a board game and several snack packs.

Given is the weight of board game = 4 pounds.

The weight of one snack pack = [tex] \frac{1}{2} \;pounds = 0.5 \;pounds [/tex]

The total weight would be less than 25 pounds.

Solving for part A :

Let 'x' represents the number of snack packs.

So the weight of all snack packs would be 0.5x pounds.

Now weight of care pack = weight of board game + weight of all snack packs.

weight of care pack = 4 + 0.5x

But total weight must be less than 25 pounds.

So the inequality would be: 4 + 0.5x < 25

Solving for part B :

Solving the inequality 4 + 0.5x < 25

⇒ 4 + 0.5x - 4 < 25 - 4

⇒ 0.5x < 21

⇒ x < 42

Solving for part C :

Since x < 42 and x is an integer number, so x = 41.

She can include maximum of 41 snack packs in the care package.

Answer:

4 batches

Step-by-step explanation:

Let

x -----> the number of cups of apples

y -----> the number of cups of mangoes

z -----> the number of cups of Blueberries

w -----> the number of cups of Blackberries

we know that

x/y=3/1 -----> equation A

x/z=3/2 ----> equation B

x/w=3/1 ----> equation C

x+w=16 ----> equation D

Solve the system of equations C and D

x=3w ----> equation C

x+w=16 ----> equation D

Substitute equation C in equation d and solve for w

3w+w=16

4w=16

w=4 cups of Blackberries

Remember that

For 1 batch of the recipe are used 1 cup of Blackberries

so

For 4 cups of Blackberries

The number of batches are

(1)*4=4 batches

Answer:

the answer is b

Step-by-step explanation:

it is

We have to prove that [tex]\sqrt{2}+\sqrt{5}[/tex] is irrational. We can prove this statement by contradiction.

Let us assume that [tex]\sqrt{2}+\sqrt{5}[/tex] is a rational number. Therefore, we can express:

[tex]a=\sqrt{2}+\sqrt{5}[/tex]

Let us represent this equation as:

[tex]a-\sqrt{2}=\sqrt{5}[/tex]

Upon squaring both the sides:

[tex](a-\sqrt{2})^{2}=(\sqrt{5})^{2}\\a^{2}+2-2\sqrt{2}a=5\\a^{2}-2\sqrt{2}a=3\\\sqrt{2}=\frac{a^{2}-3}{2a}[/tex]

Since a has been assumed to be rational, therefore, [tex]\frac{a^{2}-3}{2a}[/tex] must as well be rational.

But we know that [tex]\sqrt{2}[/tex] is irrational, therefore, from equation [tex]\sqrt{2}=\frac{a^{2}-3}{2a}[/tex] the expression [tex]\frac{a^{2}-3}{2a}[/tex] must be irrational, which contradicts with our claim.

Therefore, by contradiction,  [tex]\sqrt{2}+\sqrt{5}[/tex] is irrational.

Step-by-step explanation:

Let us assume [tex]\bf \sqrt{2}+\sqrt{5} [/tex] as rational.

[tex]\therefore \: \:\bf \sqrt{2}+\sqrt{5} = \dfrac{a}{b}[/tex]

[tex]\implies\:\:\bf \sqrt{5} = \dfrac{a}{b} - \sqrt{2} [/tex]

[tex]\implies\:\:\bf(\sqrt{5})^{2} = \bigg( \dfrac{a}{b} -\sqrt{2}\bigg)^{2} [/tex]

[tex]\implies\:\:\bf 5 = \dfrac{a^{2} }{b^{2}} - 2 \bigg(\dfrac{a}{b}\bigg) (\sqrt{2}) + (\sqrt{2})^{2}[/tex]

[tex]\implies\:\:\bf 5 = \dfrac{a^{2} }{b^{2}} - 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg) + 2[/tex]

[tex]\implies\:\:\bf 5-2 = \dfrac{a^{2}}{b^{2}}- 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)[/tex]

[tex]\implies\:\:\bf 3 = \dfrac{a^{2}}{b^{2}}- 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)[/tex]

[tex]\implies\:\:\bf 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)= \dfrac{a^{2}}{b^{2}} -3[/tex]

[tex]\implies\:\:\bf 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)=\dfrac{a^{2}-3b^{2}}{b^{2}}[/tex]

[tex]\implies\:\:\bf 2 \sqrt{2} = \bigg( \dfrac{a^{2}-3b^{2}}{b^{2}} \bigg) \bigg( \dfrac{b}{a}\bigg) [/tex]

[tex]\implies\:\:\bf \sqrt{2} =\dfrac{a^{2}-3b^{2} }{2ab}[/tex]

[tex]\textsf Hence, our assumption is wrong [/tex]

[tex] \therefore \:\:\sf {{\bf{\sqrt{2}+\sqrt{5}}} is irrational }[/tex]

[tex]\mathfrak\purple{{\purple{\bf{H}}}ence\: {\purple{\bf{P}}}roved}[/tex]

Answer:

the answer is B

Step-by-step explanation: