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3 years ago

In both of the problems where would you put parentheses to make it statement true?

Mathematics

1 answer:

Darina [25.2K]3 years ago

5 0

In the first one it would be (21-8)-6=7.
and in the second one it would be 21-(8-6)=19

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Please help me I need help with answer

Rufina [12.5K]

Answer:

+30 + -25

Savings is positive

Buying is negative

Step-by-step explanation:

We save 30 dollars

+30

We buy a game for 25

-25

+30 -25

Savings is positive

Buying is negative

7 0

3 years ago

Click the photo )) give three numbers between -7 and 9 that satisfy the given condition.

Veronika [31]

13 3/39 is too big, it's greater than 9, so that one is out.
16 is an integer, so that one is out too. 99/6 = 16.5, which is greater than 9, so it can't be right

So the answer is the other 3

6 0

3 years ago

Lannie ordered 12 copies of the same book for his book club members. The book cost $19 each, and the order has a $15 shipping ch

SashulF [63]

Hey there!

First, we need to find how much Lannie paid for all 12 copies. To do this, all we have to do is multiply the unit price ($19) by how many copies he bought (12)

$19 x 12 = $228

So, he spent $228 on the books, but we also need at add the shipping charge:

$228 + $15 = $243

Therefore, the total cost of Lannie’s order is $243

Hope this helps you!

God bless ❤️

Brainliest would be appreciated

xXxGolferGirlxXx

4 0

3 years ago

Read 2 more answers

Factor the polynomial <br>64x^3-125

kobusy [5.1K]

$64 x^{3} -125$

= $(4x)^{3} - 5^{3}$ ------> because, $(4x)^{3}$ is same as 64x³ and
$5^{3}$ is same as 125

Using the formula: $a^{3}- b^{3} =(a-b)(a ^{2} +ab+b ^{2} )$ , we get,

= $(4x -5)((4x)^{2} +4x*5+ 5^{2} )$

= $(4x -5)(16 x^{2} +20x+25)$ ------>that's your answer!!

7 0

3 years ago

F(x, y, z) = yzexzi + exzj + xyexzk,c: r(t) = (t2 + 4)i + (t2 − 1)j + (t2 − 2t)k, 0 ≤ t ≤ 2(a) find a function f such that f = ∇

exis [7]

$\nabla f(x,y,z)=\mathbf f(x,y,z)=yze^{xz}\,\mathbf i+e^{xz}\,\mathbf j+xye^{xz}\,\mathbf k$

$\dfrac{\partial f(x,y,z)}{\partial x}=yze^{xz}$
$\implies f(x,y,z)=\displaystyle\int yze^{xz}\,\mathrm dx=\dfrac{yz}ze^{xz}+g(y,z)$
$f(x,y,z)=ye^{xz}+g(y,z)$

$\dfrac{\partial f(x,y,z)}{\partial y}=e^{xz}=e^{xz}\dfrac{\partial g(y,z)}{\partial y}$
$\implies\dfrac{\partial g(y,z)}{\partial y}=0\implies g(y,z)=h(z)$
$f(x,y,z)=ye^{xz}+h(z)$

$\dfrac{\partial f(x,y,z)}{\partial z}=xye^{xz}=xye^{xz}+\dfrac{\partial h(z)}{\partial z}$
$\implies\dfrac{\partial h(z)}{\partial z}=0\implies h(z)=C$
$f(x,y,z)=ye^{xz}+C$

8 0

3 years ago