help please.
1. Oatmeal costs $1.73/lb.

How much would 2.6 lb of oatmeal cost?
(Points : 1)
$1.50
$4.48
$4.50
$4.58





Question 2. 2. What percent of $520 is $494?
(Points : 1)
5%
11%
26%
95%

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

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:

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

Answer:

(5x + 4) and (2x – 1)

Step-by-step explanation:

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]

Answer:

170.28

Step-by-step explanation:

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

Final answer:

Calculate the expected grass growth in Seattle based on the daily growth rates for sunny and cloudy days, then find the expected growth for the month of July.

Explanation:

a. Expected value of grass growth for the day:

Expected growth = (0.75 x 0.5) + (0.25 x 0.25) = 0.4375 inches

b. Expected value of grass growth for the month of July:

Expected growth for July = 31 days x Expected growth per day = 31 x 0.4375 = 13.56 inches

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

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?

Answer:

what the dude above me put he wrong

Step-by-step explanation: