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3 years ago

6

A:b= 7:1 Circle the correct equation. a= 7b b=7a a=6b b= 6a

1 answer:

astraxan [27]3 years ago

8 0

Answer:

a= 7b

Step-by-step explanation:

a=7

b=1

a->7=7b(1)

a=7(1)

a=7

7=7

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Verity's sock drawer contains 9 black socks, 5 white socks, and 6 brown socks. If Verity chooses a sock at random, what is the p

Y_Kistochka [10]

Answer:

14/20.

Step.-by-step explanation:

There are 20 socks in the drawer.

The number which are not brown = 9 + 5 = 14.

So the required probability = 14/20.

3 0

3 years ago

Which ordered pair makes both inequalities true? y < 3x – 1 y > –x + 4 On a coordinate plane, 2 straight lines are shown.

Katen [24]

Answer:

(4,0)

Step-by-step explanation:

we have

$y< 3x-1$ ----> inequality A

$y \geq -x+4$ ----> inequality B

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

Verify each ordered pair

case 1) (4,0)

<em>Inequality A</em>

$0< 3(4)-1$

$0< 11$ ----> is true

<em>Inequality B</em>

$0 \geq -(4)+4$

$0 \geq 0$ ----> is true

so

the ordered pair makes both inequalities true

case 2) (1,2)

<em>Inequality A</em>

$2< 3(1)-1$

$2< 2$ ----> is not true

so

the ordered pair not makes both inequalities true

case 3) (0,4)

<em>Inequality A</em>

$4< 3(0)-1$

$4< -1$ ----> is not true

so

the ordered pair not makes both inequalities true

case 4) (2,1)

<em>Inequality A</em>

$1< 3(2)-1$

$1< 5$ ----> is true

<em>Inequality B</em>

$1 \geq -(2)+4$

$1 \geq 2$ ----> is not true

so

the ordered pair not makes both inequalities true

5 0

3 years ago

Read 2 more answers

6 4/5 + 3 2/3<br>add and simplify

hram777 [196]

Answer:

10 3/5

Step-by-step explanation:

6 4/5 + 3 4/5

First convert to improper fraction,

34/5 + 19/5

Then add the improper fractions,

=>53/5

Then convert improper fraction to mixed fraction,

=> 53/5 = 10 3/5

5 0

3 years ago

How do you solve the compound inequality?

NikAS [45]

Step-by-step explanation:

To solve a compound inequality, first separate it into two inequalities. Determine whether the answer should be a union of sets ("or") or an intersection of sets ("and"). Then, solve both inequalities and graph.

Hopes this helps.

6 0

3 years ago

For each of the following vector fields

olga nikolaevna [1]

(A)

$\dfrac{\partial f}{\partial x}=-16x+2y$

$\implies f(x,y)=-8x^2+2xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=10y$

$\implies g(y)=5y^2+C$

$\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}$

(B)

$\dfrac{\partial f}{\partial x}=-8y$

$\implies f(x,y)=-8xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x$

$\implies \dfrac{\mathrm dg}{\mathrm dy}=x$

But we assume $g(y)$ is a function of $y$ alone, so there is not potential function here.

(C)

$\dfrac{\partial f}{\partial x}=-8\sin y$

$\implies f(x,y)=-8x\sin y+g(x,y)$

$\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=4y$

$\implies g(y)=2y^2+C$

$\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}$

For (A) and (C), we have $f(0,0)=0$ , which makes $C=0$ for both.

4 0

3 years ago