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uysha [10]
3 years ago
13

Which expression is equivalent to -4x - 36?

Mathematics
1 answer:
Julli [10]3 years ago
6 0

D. -4(x+9)

For this you distribute the number outside the parentheses to both numbers inside.

A. 4(x-9)=

4x-36

B. 2(2x-18)=

4x-36

C. -2(2x-18)=

-4x+36

D. -4(x+9)=

-4x-36

Therefore, D is the answer

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I need the answer to this please.​
ohaa [14]

Answer:

Step-by-step explanation:

Its (1,3) and (-3, -2)

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3 years ago
Two thirds of a number is no more than -10
Aliun [14]
Let x be the number in the problem. When it states that x is "no more than -10", it means than it is less than or equal to -10. So, we have

\frac{2}{3}x≤-10

In order to cancel out the \frac{2}{3} on the left and isolate x, we multiply both sides by \frac{3}{2} (since \frac{3}{2}·\frac{2}{3}=1). Thus, we have

x≤-10·\frac{3}{2}=-15

Therefore, x≤-15

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3 years ago
What is the unit form using the largest unit possible for 4tens+6tens
Gre4nikov [31]
40 + 60 = 100

Hundrends place
4 0
4 years ago
Bella's Bakery charges $2.25 per cupcake
Sveta_85 [38]

Answer:

6 cookies

Step-by-step explanation:

first, find how much money they spent on cupcakes

6 × $2.25 = $13.5

then, take $13.5 and subtract that by $24

$24 - $13.5 = $10.5

so the person spent $10.5 on cookies. divide the amount per cookie by $10.5

$10.5 ÷ $1.75 = 6

so they bought 6 cookies

8 0
3 years ago
Use the Fundamental Theorem for Line Integrals to find Z C y cos(xy)dx + (x cos(xy) − zeyz)dy − yeyzdz, where C is the curve giv
Harrizon [31]

Answer:

The Line integral is π/2.

Step-by-step explanation:

We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:

f_x = ycos(xy)

f_y = xcos(xy) - ze^{yz}

f_z = -ye^{yz}

we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)  

(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

\int {-ye^{yz}} \, dy = \int {-e^{u} \, dy} = -e^u +K = -e^{yz} + K(z)

Where, again, the constant of integration depends on Z.

As a result,

f(x,y,z) = cos(xy) - e^{yz} + K(z)

if we derivate f over z, we obtain

f_z(x,y,z) = -ye^{yz} + d/dz K(z)

That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)

The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.

3 0
4 years ago
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