All categories

2 years ago

How I can answer this question, NO LINKS, if you answer correctly I will give u brainliest!

Mathematics

1 answer:

Kazeer [188]2 years ago

5 0

Answer:

It's the first one. We notice that for every package there are 36/3 = 12 pencils. Therefore for 4 packages there will be 4*12 = 48 pencils.

You might be interested in

SUZANNE rode her bike 15mph for 12 mins then rode 12mph for 20 mins how far did she ride

Anika [276]

Answer:

7 miles

Step-by-step explanation:

Distance = Rate times Time

Distance #1: (15 mph)(12/60 hr) = 3 miles

Distance #2: (12 mph)(20/60 hr) = 4 miles

Total distance ridden = 7 miles

7 0

3 years ago

Evaluate the expression when m=3 and n=44. 1. n-6m

seraphim [82]

Answer: 35

Step-by-step explanation: PEMDAS n-6(3)

44-9

8 0

2 years ago

Read 2 more answers

Help me please don't know this

larisa86 [58]

n/5 = -1/5

5n = -5

n = -1

n - 13 = -12

n = -12 + 13

n = 1

n + 15 = -10

n = -10 - 15

n = -25

-5n = 1

n = -1/5

4 0

3 years ago

Read 2 more answers

For the differential equations dydx=sqrt(y^2−36) does the existence/uniqueness theorem guarantee that there is a solution to thi

julsineya [31]

Answer:

1. (-4,6) there is no a solution to the equation through this point

2. (2,−6) there is no a solution to the equation through this point

3. (−5,39) there is a solution to the equation through this point

4. (−1,45) there is a solution to the equation through this point

Step-by-step explanation:

Using the existence and uniqueness theorem:

Let:

$F(x,y)=\sqrt{y^2-36} \\\\and\\\\\frac{\partial F}{\partial y} =\frac{y}{\sqrt{y^2-36} }$

Now, let's find the domain of $F(x,y)$ , due to the square root:

$y^2-36 \geq 0\\\\y^2\geq 36\\\\y \geq 6\hspace{12}or\hspace{12}y\leq-6$

So the domain of the function is:

$y \in R\hspace{12}y\geq6\hspace{12}or\hspace{12}y\leq-6$

Now, due to the fraction $\frac{\partial F}{\partial y}$ the denominator must be also different from 0, so:

$y^2-36\neq0\\\\y \neq \pm6$

So, the theorem tells us that for each $y_0\in R:\hspace{12}y_0>6\hspace{12}or\hspace{12}y_0$ there exists a unique solution defined in an open interval around $x_0$ .

1. (-4,6) there is no a solution to the equation through this point because $y_0=6$

2. (2,−6) there is no a solution to the equation through this point because

$y_0=-6$

3. (−5,39) there is a solution to the equation through this point because

$y_0>6$

4. (−1,45) there is a solution to the equation through this point because

$y_0>6$

6 0

3 years ago

Find an equation of variation where y varies jointly as x and the square of z, and where y=70, when x=21 and z=5.

Bezzdna [24]

Answer:

$y = 181.75xz^2$

Step-by-step explanation:

Given

y jointly varies to x and z^2

Required

Find the equation

We start by writing out the variation;

$y\ \alpha\ xz^2$

Write as equation

$y = kxz^2$

Substitute values of x,y and z

$70 = k * 21 * 25^2$

$70 = k * 21 * 25 * 25$

$70 = k * 13125$

Divide both sides by 7

$\frac{70}{13125} = \frac{k * 131251}{13125}$

$\frac{70}{13125} = k$

$187.5 = k$

$k = 181.75$

Substitute k in $y = kxz^2$

$y = 181.75xz^2$

4 0

3 years ago