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3 years ago

What expression is equivalent to x^2-7x+12+(x-3)^2

Mathematics

1 answer:

nikitadnepr [17]3 years ago

3 0

Answer:

2x²−13x+21

Step-by-step explanation:

Hope this answer helps you :)

Have a great day

Mark brainliest

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Approximately when where there 500 gallons of water in the pool

maw [93]

look it up

Step-by-step explanation:

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2 years ago

Jenny is saving $18 a week. She wants to buy a cell

tester [92]

14
Because you divide

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2 years ago

Read 2 more answers

Can someone please explain to me how to get this answer?

SIZIF [17.4K]

Answers:
A = 120
b = 45.0
c = 33.2

Side Note: only one triangle is possible
See attached for a visual. I used GeoGebra to draw the triangle.

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Explanation:

We are given the following information
B = 35
C = 25
a = 68

We need to find the following
A, b, c

where the lower case letters represent the side lengths; the upper case letters are the angles. The angles are opposite their corresponding sides. For instance, side lowercase b is opposite angle uppercase B. The other letters are positioned the same way.

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First use the idea that for any triangle, the three angles (A,B,C) must add to 180 degrees
A+B+C = 180
Replace B and C with 35 and 25. Solve for angle A
A+35+25 = 180
A+60 = 180
A+60-60 = 180-60
A = 120

Now that we know the three angles A = 120, B = 35, C = 25, we can find the missing sides 'b' and 'c'

-----------------------

We will use the law of sines to find side b
sin(A)/a = sin(B)/b
sin(120)/68 = sin(35)/b
b*sin(120) = 68*sin(35) <<--- cross multiply
b = 68*sin(35)/sin(120)
b = 45.037013350222 <<--- use a calculator; this value is approximate
b = 45.0 <<--- round to the nearest tenth

-----------------------

Do the same for side c
sin(A)/a = sin(C)/c
sin(120)/68 = sin(25)/c
c*sin(120) = 68*sin(25) <<--- cross multiply
c = 68*sin(25)/sin(120)
c = 33.1838323365404 <<--- use a calculator; this value is approximate
c = 33.2 <<--- round to the nearest tenth

4 0

3 years ago

Read 2 more answers

Darren wins a coupon for $4 off the lunch special for each of five days he pays $75 for his 5 lunch specials write and solve an

Valentin [98]

The original price for one lunch special is $19.

Explanation

The original price for one lunch special is 'p' dollar.

He wins a coupon for $4 off for each of five days. That means , he needs to pay $(p-4)$ dollar each day.

So, the total amount needed to pay for 5 days $= 5(p-4)$ dollar

Given that, he pays $75 for his 5 lunch specials. So the equation will be.....

$5(p-4)= 75\\ \\ 5p-20=75\\ \\ 5p= 75+20\\ \\ 5p= 95\\ \\ p=\frac{95}{5}=19$

So, the original price for one lunch special is $19.

8 0

2 years ago

(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim

aleksklad [387]

(i) Given that

$V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr$

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

$V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}$

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

5 0

2 years ago