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3 years ago

One of the solutions to x2 − 2x − 15 = 0 is x = −3. What is the other solution?

2 answers:

7 0

If you factor the equation you can see what the solutions are. Two factor a quadratic of the form ax^2+bx+c, find two values which satisfy two conditions...

jk=ac=-15 and j+k=b=-2 so j and k must be -5 and 3 so the factors are:

(x-5)(x+3)

So the other solution is x=5

Tcecarenko [31]3 years ago

6 0

The answer is D) X=5 on E2020 (:

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A cylinder has a volume of 390 cm³. If a cone has the same dimensions as the cylinder then what is the volume of the cone.

dangina [55]

Answer: The volume of the cone is 130 cubic centimeters.

Step-by-step explanation:

We know that to find the volume of a cylinder you will use the formula

V= $\pi$ $r^{2} *h$ Where r is the radius and h is the height.

And to find the volume of a cone you will use the formula

V= $\pi$ $r^{2}$ *h /3

So if they have the same dimensions which is the height and the radius then we will just divide the volume of the cylinder by 3 to find the volume of the cone.

V = 390/3

V = 130 cm^3

8 0

3 years ago

Macy's if offering a sale where you can use two different coupons, Coupon A gives 15%

Juliette [100K]

The ticket price of the item after applying coupon B first and then coupon A is $65.45

Coupons reduce the ticket price of an item.

Ticket price after coupon B is applied

An item costs $80. After applying the coupon B which gives, $3 off the price, the cost of the item reduces to $77 ($80 - $3).

The ticket price of the item after coupon B is applied is $77

Ticket price after coupon A is applied

The item costs $77 when coupon B is applied. If the discount is 15% off on coupon A, it means that the item costs 85%( 100 - 15%) of its initial price.

Ticket Price of the item = percentage price x price of the item after the first coupon B was applied

85% x $77

0.85 x $77 = $65.45

A similar question was solved here: brainly.com/question/17413216?referrer=searchResults

8 0

2 years ago

the half-life of strontium-90 is approximately 29 years. how much of a 500 g sample of strontium-90 will remain after 58 years

Yuliya22 [10]

Answer: 125 g

Step-by-step explanation:

$A = P_o\cdot e^{kt}\\\\\text{First, use the given information to find k:}\\\\\bullet A=\dfrac{1}{2}P_o\\\\\bullet k = unknown\\\\\bullet t=29\text{ years}\\\\\dfrac{1}{2}P_o=P_o\cdot e^{k(29)}\\\\\\\dfrac{1}{2}=e^{k(29)}\qquad divided\ both\ sides\ by\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=ln\bigg(e^{k(29)}\bigg)\qquad applied\ ln\ to\ both\ sides\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=29k\qquad simplified-ln\ and\ e\ cancel\ out\\\\\\\dfrac{ln\bigg(\dfrac{1}{2}\bigg)}{29}=k\qquad divided\ 29\ from\ both\ sides\\\\\\-0.0239=k$

$\text{Now, use the following in the equation to solve for A:}\\\\\bullet A=unknown\\\bullet P_o=500\\\bullet k=-0.0239\\\bullet t=58\text{ years}\\\\A=500\cdot e^{(-0.0239)(58)}\\\\.\quad=500\cdot e^{-1.386}\\\\.\quad=125$

8 0

3 years ago

How to divide 33 photos into two groups so the ratio is 4 to 7

Nataliya [291]

33 photos in a 4x to 7x ratio

4x + 7x = 33
11x = 33
x = 3

33 photos can be put in a 4(3) to 7(3) ratio

which is a 12 to 21 ratio

so 12 photos on one side, and 21 on the other side to get the ratio of 4 to 7

Note:
a(b) = ab = a x b = a times b

3 0

3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let X = number of students who read above the eighth grade level.

(a)

A sample of n = 269 students are selected. Of these 269 students, X = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

$\hat p=\frac{X}{n}=\frac{224}{269}=0.8327$

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

$1-\hat p=1-0.8327$

$=0.1673$

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

n = 709

X = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

$\hat q=1-\hat p$

$=1-\frac{X}{n}$

$=1-\frac{546}{709}$

$=0.2299\\\approx 0.229$

The critical value of z for 95% confidence interval is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for the population proportion as follows:

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)$

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0

3 years ago