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WITCHER [35]
3 years ago
6

The sum of two numbers is 12 and their difference is 4. What is the product of these two numbers

Mathematics
2 answers:
Tju [1.3M]3 years ago
8 0

Answer:

the answer is 32

Step-by-step explanation:

12-4=8

add 8 and 4 you'll get 12. Subtract 4 from 8 you'll get 4.

multiply 8 and 4 you'll get 32

shutvik [7]3 years ago
7 0

Answer:

48

Step-by-step explanation:

because I had times 12 and 4 to get 48

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Let f(x)=x^2 and g(x)=x-3. evaluate (fog)(-2)
Studentka2010 [4]
F(x) = x²
g(x) = x - 3

g(-2) = (-2) - 3 = -5

f(-5) = (-5)² = 25

Answer : <span>(fog)(-2) = 25</span>
7 0
3 years ago
Let f(x) = - x ^ 3 - (4x - 1) Find f(- 5)
timofeeve [1]

Answer:

146

Step-by-step explanation:

replace every x with -5

3 0
3 years ago
How to solve this pythagoras theorem
My name is Ann [436]

Answer:

see explanation

Step-by-step explanation:

The hypotenuse is the longest side thus is (4x + 1)

The legs are 2x and (4x - 1)

Using Pythagoras' theorem, then

(4x + 1)² = (2x)² + (4x - 1)² ← expanding factors

16x² + 8x + 1 = 4x²  + 16x² - 8x + 1 , that is

16x² + 8x + 1 = 20x² - 8x + 1 ( subtract 20x² - 8x + 1 from both sides )

- 4x² + 16x = 0 ( multiply through by - 1 )

4x² - 16x = 0 ← factor out 4x from each term

4x(x - 4) = 0

Equate each factor to zero and solve for x

4x = 0 ⇒ x = 0

x - 4 = 0 ⇒ x = 4

Now x > 0, thus x = 4

2x = 2(4) = 8

4x - 1 = 4(4) - 1 = 16 - 1 = 15

4x + 1 = 4(4) + 1 = 16 + 1 = 17

Thus

perimeter = 8 + 15 + 17 = 40 cm

3 0
3 years ago
Helppp im not sure about this one
PSYCHO15rus [73]
There’s no question here!
8 0
3 years ago
Evaluate the line integral, where C is the given curve. (x + 6y) dx + x2 dy, C C consists of line segments from (0, 0) to (6, 1)
Dima020 [189]

Split C into two component segments, C_1 and C_2, parameterized by

\mathbf r_1(t)=(1-t)(0,0)+t(6,1)=(6t,t)

\mathbf r_2(t)=(1-t)(6,1)+t(7,0)=(6+t,1-t)

respectively, with 0\le t\le1, where \mathbf r_i(t)=(x(t),y(t)).

We have

\mathrm d\mathbf r_1=(6,1)\,\mathrm dt

\mathrm d\mathbf r_2=(1,-1)\,\mathrm dt

where \mathrm d\mathbf r_i=\left(\dfrac{\mathrm dx}{\mathrm dt},\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt

so the line integral becomes

\displaystyle\int_C(x+6y)\,\mathrm dx+x^2\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}(x+6y,x^2)\cdot(\mathrm dx,\mathrm dy)

=\displaystyle\int_0^1(6t+6t,(6t)^2)\cdot(6,1)\,\mathrm dt+\int_0^1((6+t)+6(1-t),(6+t)^2)\cdot(1,-1)\,\mathrm dt

=\displaystyle\int_0^1(35t^2+55t-24)\,\mathrm dt=\frac{91}6

6 0
3 years ago
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