Dyann takes a bus to work every day. On Monday the trip took 34 minutes. On Tuesday the trip took 36 minutes. On Wednesday the trip took 35 minutes. On Thursday the trip took 34 minutes. Dyann determines that the trip to work on Friday will take about 35 minutes.

Does her conclusion use inductive, or deductive reasoning?

The student's question pertains to solving a system of equations to find two numbers that multiply to a given product and add to a given sum. The solution involves testing pairs of factors of the product until the pair that also sums to the required number is found, for example, finding that 3 and 8 multiply to 24 and add to 11.

The student is looking to solve a classic problem in algebra: finding two numbers that, when multiplied together, result in a given number (the product), and when added together, sum to another given number (the sum). This is often related to factoring a quadratic equation or finding the roots of a polynomial. The solution involves setting up a system of equations based on the given conditions and solving for the two unknown numbers. This process typically requires identifying the correct pair of numbers that satisfy both the multiplication and addition conditions. Let's take an example:

Suppose we need to find two numbers that multiply to make 24 and add to make 11. We can denote these numbers as x and y, leading to the following equations:

By systematically testing pairs of factors of 24 (e.g., 1 and 24, 2 and 12, 3 and 8, 4 and 6), we find that 3 and 8 satisfy both conditions, because 3 . 8 = 24 and 3 + 8 = 11. Therefore, the two numbers are 3 and 8.

Answer: [tex](-1,\frac{1}{3}),\ (0,1),\ (1,3),\ (2,9),\ (3,27),\ (4,81)[/tex]

Step-by-step explanation:

The given exponential function :

[tex]y=3^x[/tex]

At x = -1, we get

[tex]y=3^{-1}=\dfrac{1}{3}[/tex]

At  x = 0, we get

[tex]y=3^{0}=1[/tex]

At x = 1, we get

[tex]y=3^{1}=3[/tex]

At x = 2, we get

[tex]y=3^{2}=9[/tex]

At x = 3, we get

[tex]y=3^{3}=27[/tex]

At x = 4, we get

[tex]y=3^{4}=81[/tex]

Hence, the completed ordered pairs are : [tex](-1,\frac{1}{3}),\ (0,1),\ (1,3),\ (2,9),\ (3,27),\ (4,81)[/tex]

Ques 1)

Option: C is the correct answer.

c. 9 inches square.

Ques 2)

Option: c

c. By simplifying 8 to 2^3 to make both powers base two and subtracting the exponents.

Ques 3)

Option: b

b. 2 to the 7 over 3 power

Ques 1)

A rectangle has a length of the cube root of 81 inches and a width of 3 to the 2 over 3 power inches.

i.e. let 'l' and 'b' denote the length and width of the rectangle.

i.e.[tex]l=\sqrt[3]{81}\\\\l=(3^4)^{\dfrac{1}{3}}\\\\l=3^{\dfrac{4}{3}[/tex]

since,

[tex](a^m)^n=a^{mn}[/tex]

and

[tex]w=3^{\dfrac{2}{3}}[/tex]

Hence, the area of rectangle is given by:

[tex]Area=l\times w\\\\\\Area=3^{\dfrac{1}{3}}\times3^{\dfrac{4}{3}}\\\\\\Area=3^({\dfrac{1}{3}+\dfrac{4}{3}})\\\\\\Area=3^{\dfrac{6}{3}}\\\\\\Area=3^2\\\\\\Area=9\ square\ inches[/tex]

As we know that:

[tex]a^m\times a^n=a^{m+n}[/tex]

Hence, Area=9 square inches.

Ques 2)

2 to the fifth power, over 8 = 2 to the 2nd power

i.e. we need to prove that:

[tex]\dfrac{2^5}{8}=2^2[/tex]

As we know that:

[tex]\dfrac{2^5}{8}=\dfrac{2^5}{2^3}=2^{5-3}=2^2[/tex]

( since

[tex]\dfrac{a^m}{a^n}=a^{m-n}[/tex]  )

Hence, the correct answer is: Option: c

Ques 3)

The cube root of 2 to the seventh power.

i.e.

[tex]\sqrt[3]{2^7}\\\\=(2^7)^{\dfrac{1}{3}}\\\\=2^{\dfrac{7}{3}}[/tex]

Since,

[tex]\sqrt[n]{x}=x^{\dfrac{1}{n}}[/tex]

and

[tex](a^m)^n=a^{mn}[/tex]

Hence, the correct answer is: option: b

1a.

Answer is 8x² - 2x + 8

  (x² - 4x + 5) + (7x² + 2x + 3)

The parentheses can go away since this is all addition (associative property of addition)

  x² - 4 x + 5 + 7x² + 2x + 3

Combine like terms.

  (x^2 + 7x²) + (-4x + 2x) + (5 + 3)   = 8x² - 2x + 8

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1b.

Answer is 5x^2 + 7x - 7

  (7x² + 4x - 6) - (2x² - 3x + 1)

Distribute the negative into the right parentheses.

  (7x² + 4x - 6) - (2x² - 3x + 1)

  = 7x² + 4x - 6 - 2x² + 3x - 1

  = (7x² - 2x²) + (4x + 3x) + (-6 - 1)

  =  5x^2 + 7x - 7

Answer:

The height of dam =45.5 m.

Step-by-step explanation:

We are given that Scarlett stands 90 m away from the dam and records the angle of elevation to the top of the dam to be [tex]26^{\circ}p[/tex]

Scarelett's height is 1.65 meters.

We have to find the height of the dam.

Let h be the height of dam

AC=AB+BC

BC=x

h=1.65+x

CD=EB=90 m

In triangle ABE

[tex]\theta=26^{\circ}[/tex]

[tex]tan\theta=\frac{perpendicular\;side}{Base}[/tex]

[tex]tan26^{\circ}=\frac{AB}{90}[/tex]

[tex]0.4877=\frac{x}{90}[/tex]

[tex]x=0.4877\times 90[/tex]

[tex]x=43.893 m[/tex]

Therefore, the height of dam=1.65+43.893=45.543 m

Answer: The height of dam =45.5 m

Andrew's price per puppy treat, after using a $2.80 coupon on a $9.94 bag that contains 42 treats, will be $0.17.

Andrew has a coupon that will reduce the price of puppy treats. Without the coupon, one bag of treats costs $9.94 and contains 42 treats. To find the price per treat after using the $2.80 coupon, we first subtract the coupon value from the original price of the bag. This gives us the discounted price of the bag:

Discounted Price = Original Price - Coupon Value = $9.94 - $2.80 = $7.14

Next, we find the price per treat by dividing the discounted price of the bag by the number of treats in the bag:

Price per Treat = Discounted Price ÷ Number of Treats = $7.14 ÷ 42 = $0.17 (rounded to two decimal places)

After using the coupon, the price per puppy treat will be $0.17.

Answer:

170.28

Step-by-step explanation:

I just took the test and am reviewing the correct answers.

y = 6x2, y = −4.5x2, y = −x2

This should be correct answer:) He got it mixed up in his answer below when graphing.:)

For a better understanding of the solution and the explanation given here please go through the diagram in the file attached.

By definition, measure of an arc is the angle subtended by the arc at the centre of the circle. Thus, the measure of the arc [tex] \overarc {RSQ} [/tex] is the measure of the [tex] \angle RAQ [/tex] (in red) which the arc [tex] \overarc {RSQ} [/tex] subtends at the center.

Now, in order to find the measure of the [tex] \angle RAQ [/tex] (in red) we will have to use the information given in the question. We have been told that [tex] m\angle RSQ=24^{\circ} [/tex]. Now, we know that the double angle theorem says that, the angle subtended at the centre of a circle is double the size of the angle subtended at the circumference from the same two points. Therefore, applying this theorem we get:

[tex] \angle RAQ =2\times 24^{\circ}=48^{\circ} [/tex]

Thus if [tex] \angle RAQ [/tex] (in red)=[tex] 48^{\circ} [/tex], then [tex] \angle RAQ [/tex] (in green) will be equal to measure of the arc [tex] \overarc {RSQ} [/tex]. Now, [tex] \angle RAQ [/tex] (in green) will obviously be:

[tex] \angle RAQ [/tex] (in green)=[tex] 360^{\circ}-\angle RAQ [/tex] (in red)

[tex] \angle RAQ [/tex] (in green)=[tex] 360^{\circ}-48^{\circ} [/tex]

[tex] \angle RAQ [/tex] (in green)=[tex] 312^{\circ} [/tex]

Thus, the measure of arc [tex] \overarc {RSQ} [/tex]=[tex] 312^{\circ} [/tex]

Answer:

what the dude above me put he wrong

Step-by-step explanation:

Answer:  The correct option is (d) five over six.

Step-by-step explanation:  Given that John rolls a number cube twice. We are to find the probability that the sum of the 2 rolls is less than 8 given that he first roll is a 1.

The sample space of an event of rolling a cube is

S = {1, 2, 3, 4, 5, 6}.

That is, n(S) = 6.

Now, let 'A' be the event that the sum of the two rolls is less than 7, then

A = {1, 2, 3, 4, 5}.

That is, n(A) = 5.

So, the probability of happening of event A is given by

[tex]P(A)=\dfrac{n(A)}{n(S)}=\dfrac{5}{6}.[/tex]

Thus, the required probability is five over six.

Option (d) is correct.

To find the zeros of the quadratic function [tex]\( f(x) = 8x^2 - 16x - 15 \)[/tex], you need to solve for [tex]\( x \) when \( f(x) = 0 \)[/tex].

So, you set \( f(x) \) equal to zero:

[tex]\[ 8x^2 - 16x - 15 = 0 \][/tex]

To solve this quadratic equation, you can use the quadratic formula:

[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]where \( a = 8 \), \( b = -16 \), and \( c = -15 \).[/tex]

Plugging these values into the quadratic formula:

[tex]\[ x = \frac{{-(-16) \pm \sqrt{{(-16)^2 - 4(8)(-15)}}}}{{2(8)}} \]\[ x = \frac{{16 \pm \sqrt{{256 + 480}}}}{{16}} \]\[ x = \frac{{16 \pm \sqrt{{736}}}}{{16}} \][/tex]

Now, you simplify the expression under the square root:

[tex]\[ \sqrt{736} = \sqrt{16 \times 46} = 4\sqrt{46} \][/tex]

So, the expression becomes:

[tex]\[ x = \frac{{16 \pm 4\sqrt{46}}}{{16}} \][/tex]

Now, you can simplify further:

[tex]\[ x = \frac{{4(4 \pm \sqrt{46})}}{{4 \times 4}} \]\[ x = \frac{{4 \pm \sqrt{46}}}{{4}} \][/tex]

This gives you two solutions:

[tex]\[ x = \frac{{4 + \sqrt{46}}}{{4}} \]\[ x = \frac{{4 - \sqrt{46}}}{{4}} \][/tex]

These are the zeros of the quadratic function [tex]\( f(x) = 8x^2 - 16x - 15 \).[/tex]

Complete question :

What are the zeros of the quadratic function f(x) = 8x2 - 16x - 15?