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3 years ago

How do I do this question!

Mathematics

2 answers:

irina1246 [14]3 years ago

4 0

Answer: $\frac{7}{10}$

Step-by-step explanation:

2 $\frac{1}{10}$ barrels of almonds to make 3 batches of granola bar.

We need to turn fraction 2 $\frac{1}{10}$ into a mixed number

2 $\frac{1}{10}$ → $\frac{21}{10}$

Now we set up an equation.

$\frac{\frac{21}{10} }{3} =\frac{x}{1}$

Solve for x by cross multiplying and dividing.

$\frac{21}{10}$ ÷ 3 (KCF)

$\frac{21}{10}$ × $\frac{1}{3}$ = $\frac{21}{30}$

Now we simplify the fraction.

$\frac{21}{30}$ → $\frac{7}{10}$

$\frac{7}{10}$ barrels of almonds were used in each batch.

Flauer [41]3 years ago

3 0

Answer:2 x 3

Step-by-step explanation:

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Part A: The club has learned from experience over the years that when they do not charge an admission fee, attendance is 800 peo

sesenic [268]

Answer:

750

Step-by-step explanation:

800 - 8p = 750

5 0

3 years ago

HELP ASAP IM GIVING 40 POINTS I NEED THE AREA FROM THIS TRAPEZOID MODEL IN PICTURE

Nitella [24]

Answer:

51/2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Y = a x 3 + b x 2 + c x + d

Alla [95]

Answer:

If its a cube function. Then I think this is the answer. Can you mark brainliest? And can you tell me if it's right or wrong?

Step-by-step explanation:

Consider the cubic function f(x) = ax^3 + bx^2 + cx + d. Determine the values of the constants a, b, c and d so that f(x) has a point of inflection at the origin and a local maximum at the point (2,...

Consider the cubic function f(x) = ax^3 + bx^2 + cx + d. Determine the values of the constants a, b, c and d

so that f(x) has a point of inflection at the origin and a local maximum at the point (2, 4). ==>c=3

8 0

3 years ago

A one-parameter family of solutions of the de p' = p(1 − p) is given below.

tensa zangetsu [6.8K]

Answer:

A solution curve pass through the point (0,4) when $c_{1} = -\frac{4}{3}$ .

There is not a solution curve passing through the point(0,1).

Step-by-step explanation:

We have the following solution:

$P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}$

Does any solution curve pass through the point (0, 4)?

We have to see if P = 4 when t = 0.

$P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}$

$4 = \frac{c_{1}}{1 + c_{1}}$

$4 + 4c_{1} = c_{1}$

$c_{1} = -\frac{4}{3}$

A solution curve pass through the point (0,4) when $c_{1} = -\frac{4}{3}$ .

Through the point (0, 1)?

Same thing as above

$P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}$

$1 = \frac{c_{1}}{1 + c_{1}}$

$1 + c_{1} = c_{1}$

$0c_{1} = 1$

No solution.

So there is not a solution curve passing through the point(0,1).

3 0

3 years ago

Is it considered a parallelogram?

scZoUnD [109]

C

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6 0

3 years ago