Find the X-intercepts : 0=2(x+2)^2-5

What 2+2 x10+2X2+10=

erastova [34]

Answer:

36

Step-by-step explanation:

Use PEMDAS:

2*10 is 20

2*2 is 4

So the equation is now

2+20+4+10=

Now add:

36

Thus the solution is 36

Hope this helps!

4 0

2 years ago

Read 2 more answers

PLS SOLVE ASAP!!! These points are linear. Find the slope.

Lubov Fominskaja [6]

Could be y=8x + 16?
The gradient is 8 bcause each time x increases by one y increases by 8

5 0

2 years ago

10 gallons of gasoline costs $30. The cost of gasoline per gallon is ________ and gallons of gasoline per dollar are ________.

Blababa [14]

2. if the cost of 10 gallons of gasoline is 30 dollars then divide 10 by 10 and divide 30 by 10 to get 1 and 3 so the cost of 1 gallon is $3

3. if 2 packets of bread is 3 dollars then divide 2 and 3 by 2 so in conclusion 1 loaf f bread is $1.50.

4. if a cleaning service charges $1000 for 20 offices then divide 1000 by 20 and 20 by 20 to have an answer of $50 for every one office.

5. to find the answer to this problem you need to divide 320 by 4 to find out organes per bag so 80 oranges per bag.

6. if the baker ca bake 12 cakes every 45 minutes then he can bake 4 cakes every 15 minutes because divide both 12 and 45 by three so in conclusion the baker can bake 16 cakes every hour.

7. so if you can print 100 pages for $2 then multiply 100 and 2 by two to get 200 pages for $4

8. so to find the answer to this all you need to do is divide 12 by six to fin out that it is 2 apples for $1 and for the other store you divide 21 by seven to find out that it is 3 apples for $1.

9. to find out the answer for this divide 24,000 by 12 to find that his monthly income is $2,000.

10. if a 0.2 pund package is $4 then a 0.001 pound package is 20 cents.

I hope i got the answers right and that i could help! :)

6 0

2 years ago

In ΔUVW, w = 44 cm, u = 83 cm and ∠V=141°. Find the area of ΔUVW, to the nearest square centimeter.

Tema [17]

Answer:

$Area \approx 1149\ cm^2$

Step-by-step explanation:

Given that:

ΔUVW,

Side w = 44 cm, (It is the side opposite to $\angle W$ )

Side u = 83 cm (It is the side opposite to $\angle U$ )

and ∠V=141°

Please refer to the attached image with labeling of the triangle with the dimensions given.

Area of a triangle with two sides given and angle between the two sides can be formulated as:

$A = \dfrac{1}{2}\times a\times b\times sinC$

Where a and b are the two sides and

$\angle C$ is the angle between the sides a and b

Here we have a = w = 44cm

b = u = 44cm

and ∠C= ∠V=141

Putting the values to find the area:

$A = \dfrac{1}{2}\times 44\times 83\times sin141\\\Rightarrow A = \dfrac{1}{2} \times 3652 \times sin141\\\Rightarrow A =1826 \times 0.629\\\Rightarrow A \approx 1149\ cm^2$

So, the area of given triangle to the nearest square centimetre is:

$Area \approx 1149\ cm^2$

8 0

3 years ago

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f

Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen who carry a credit card balance.

The formula for the interval is:

p'₁± $Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }$

$Z_{1-\alpha /2}= Z_{0.95}= 1.648$

0.37±1.648* $\sqrt{\frac{0.37*0.63}{1000} }$

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂± $Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }$

0.48±1.648* $\sqrt{\frac{0.48*0.52}{1000} }$

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: $\sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015$

Seniors: $\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016$

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0

3 years ago