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3 years ago

Write two expressions where the solution is 19.

Mathematics

2 answers:

sergij07 [2.7K]3 years ago

7 0

S-1=18
x/10=1.9
these have solution 19

liberstina [14]3 years ago

7 0

Answer:10+9=19. 75÷3=19

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Please assist me with this problem

Anna11 [10]

Answer:

Option B is not a justification

Step-by-step explanation:

Option A is a justification (to get two pair of similar triangles (ABC-ACD) and (ABC-BCD))

Option C is a justification (a^2 = cy, b^2 = cx, a^2 + b^2 = cy + cx)

Option D is a justification (a^2 + b^2 = cy + cx = c(y + x) = c^2)

=> Option B is not a justification

8 0

3 years ago

Malik earned d dollars raking leaves. His friend, Isaiah, earned

erastova [34]

Answer:

d=3d-5 or the other way around

p.s mark me as brainlist if correct

3 0

3 years ago

Please help with the question 2 to 18

vladimir1956 [14]

Since these questions all require similar methods of solving, I will only completely work out the first question of each type.

2. To find $\frac{7}{10}$ as a decimal, all we need to do is complete the fraction by dividing the numerator by the denominator.

$\frac{7}{10}$ = 7 / 10 = 0.7

$\frac{7}{10}$ as a decimal is 0.7.

3. To write a fraction as a percentage, we must first convert it into a decimal, the way we did in problem 2.

$\frac{13}{20} =0.65$

Then to change that number into a percentage, we can either multiply it by 100, or (my personal shortcut) just shift the decimal point over two places to the right.

0.65 = 65%

which is the answer to question 3.
$\frac{13}{20}$ written as a percent is 65%.

4. Writing a percentage as a fraction is a bit easier than writing a fraction as a percentage. All you need to do is place the percentage in the numerator's place of a fraction with 100 as the denominator and simplify.

2%5 = $\frac{25}{100}$

Since 25 goes into 100 four times, we can simplify this fraction.

$\frac{25}{100} = \frac{1}{4}$

which is our answer.

25% written as a fraction is $\frac{1}{4}$ .

5. This is the same as problem 3, so I won't show my work for the sake of time.

$\frac{2}{5}$ as a percentage is 40%.

6. This is the same as problem 1, so I will not show steps again.

$\frac{3}{20}$ as a decimal is 0.15.

7. Done similar problems in past. No work shown.

$\frac{21}{50}$ as a percent is 42%.

8. Same as last problem.

$\frac{1}{25}$ as a percent is 4%.

9. Same as number 4.

6% as a fraction is $\frac{3}{50}$ .

10. Similar in past.

$\frac{3}{5}$ as a percentage is 60%.

11. Same as 6.

$\frac{12}{25}$ as a decimal is 0.48.

12. Done a lot of these so far.

$\frac{3}{10}$ as a percentage is 30%.

13. Just did one like this.

$\frac{3}{4}$ as a percentage is 75%.

14. Same as 4 and 9.

65% as a fraction is $\frac{13}{20}$ .

15. Plenty of these done in past.

$\frac{1}{5}$ as a percentage is 20%.

16. Last problem, and it's just like the one just before this.

$\frac{9}{10}$ written as a percent is 90%.

So there are the results.
Hope that helped! =)

6 0

3 years ago

Read 2 more answers

Let and be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.

elena-14-01-66 [18.8K]

The given identities are verified by using operations of the del operator such as divergence and curl of the given vectors.

<h3>What are the divergence and curl of a vector field?</h3>

The del operator is used for finding the divergence and the curl of a vector field.

The del operator is given by

$\nabla=\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}$

Consider a vector field $F=x\^i+y\^j+z\^k$

Then the divergence of the vector F is,

div F = $\nabla.F$ = $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(x\^i+y\^j+z\^k)$

and the curl of the vector F is,

curl F = $\nabla\times F$ = $\^i(\frac{\partial Fz}{\partial y}- \frac{\partial Fy}{\partial z})+\^j(\frac{\partial Fx}{\partial z}-\frac{\partial Fz}{\partial x})+\^k(\frac{\partial Fy}{\partial x}-\frac{\partial Fx}{\partial y})$

<h3>Calculation:</h3>

The given vector fields are:

$F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$

1) Verifying the identity: $\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

Consider L.H.S

⇒ $\nabla.(aF1+bF2)$

⇒ $\nabla.(a(M\^i + N\^j + P\^k) + b(Q\^i + R\^j + S\^k))$

⇒ $\nabla.((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the dot product between these two vectors,

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(1)

Consider R.H.S

⇒ $a\nabla.F1+b\nabla.F2$

So,

$\nabla.F1=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(M\^i + N\^j + P\^k)$

⇒ $\nabla.F1=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}$

$\nabla.F2=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(Q\^i + R\^j + S\^k)$

⇒ $\nabla.F1=\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z}$

Then,

$a\nabla.F1+b\nabla.F2=a(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z})+b(\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z})$

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(2)

From (1) and (2),

$\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

2) Verifying the identity: $\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

Consider L.H.S

⇒ $\nabla\times(aF1+bF2)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times(a(M\^i+N\^j+P\^k)+b(Q\^i+R\^j+S\^k))$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times ((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the cross product,

$\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y})$ ...(3)

Consider R.H.S,

⇒ $a\nabla\times F1+b\nabla\times F2$

So,

$a\nabla\times F1=a(\nabla\times (M\^i+N\^j+P\^k))$

⇒ $\^i(\frac{\partial aP\^k}{\partial y}- \frac{\partial aN\^j}{\partial z})+\^j(\frac{\partial aM\^i}{\partial z}-\frac{\partial aP\^k}{\partial x})+\^k(\frac{\partial aN\^j}{\partial x}-\frac{\partial aM\^i}{\partial y})$

$a\nabla\times F2=b(\nabla\times (Q\^i+R\^j+S\^k))$

⇒ $\^i(\frac{\partial bS\^k}{\partial y}- \frac{\partial bR\^j}{\partial z})+\^j(\frac{\partial bQ\^i}{\partial z}-\frac{\partial bS\^k}{\partial x})+\^k(\frac{\partial bR\^j}{\partial x}-\frac{\partial bQ\^i}{\partial y})$

Then,

$a\nabla\times F1+b\nabla\times F2$ =

...(4)

Thus, from (3) and (4),

$\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

Learn more about divergence and curl of a vector field here:

brainly.com/question/4608972

#SPJ4

Disclaimer: The given question on the portal is incomplete.

Question: Let $F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$ be differential vector fields and let a and b arbitrary real constants. Verify the following identities.

$1)\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2\\2)\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

8 0

2 years ago

A horse ran 800 meters in 40 seconds , 1200 meters in 60 seconds , and 480 meters in 24 seconds . Is this a proportional relatio

frosja888 [35]

Answer:

This is a proportional relationship, the constant of proportionality is 20m/s and it represents that the horse can run 20 meters every second.

Equation: d = 20s, where d=distance and s=number of seconds.

Step-by-step explanation:

In order to find out whether this relationship is proportional, you need to see if the rate at which the horse runs is constant (the same). If you look at the three sets of data (24, 480), (40, 800) and (60, 1200) where the pair is (seconds, meters), you can see that for any two sets of data the change in meters divided by the change in seconds is consistently 20m/s. For example:

$\frac{1200-800}{60-40}=\frac{400}{20}=20$

Since the constant is 20, we know that the horse can run 20 meters every second. To find the horse's total distance, we need to multiply the rate by the number of seconds that it runs:

d = 20s

8 0

3 years ago