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3 years ago

5x − 3y = −11 2x − 6y = −14

Mathematics

1 answer:

Digiron [165]3 years ago

7 0

The answer is 2x-6y=-14 is the aswer

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There once was a

r-ruslan [8.4K]

I am pretty sure the only way to do this is:

• To bring the corn first

• To bring the goose second

• To bring the dog third

Hope this helped!

3 0

4 years ago

Read 2 more answers

Find the maximum rate of change of f at the given point and the direction in which it occurs. f(x, y) = 2 sin(xy), (0, 5)

andrew-mc [135]

Answer:

Step-by-step explanation:

Given the function f(x, y) = 2 sin(xy) at (0,5), the maximum rate of change of the function at that point will occur in the direction $\nabla f(0,5)$

$\nabla f(x, y) = \frac{\delta(2sin(xy)) }{\delta x} i + \frac{\delta(2sin(xy)) }{\delta y} j\\\\\nabla f(x, y) = 2ycos(xy)i + 2xcos(xy)j\\\\\nabla f(0, 5) = 2(5)cos(0*5)i + 2(0)cos(0*5)j\\\\\nabla f(0, 5) = 10cos0i+ 0j\\\\\nabla f(0, 5) = 10i+0j\\\\||\nabla f(0, 5)|| = \sqrt{10^2+0^2} \\\\||\nabla f(0, 5)|| = \sqrt{100} = 10$

Hence, the magnitude of the maximum rate of the function at the point (0, 5) is 10

6 0

3 years ago

Anne can make 6 headbands 3/5 from meters of cloth. Which expression shows the amount of cloth needed to make 1 headband? What i

Marina CMI [18]

The amount of cloth needed to make 1 headband is 10 meters.

<h3>Unit value</h3>

Number of headbands Anne made = 6

Total clothes used = 3/5 meters

Cloth needed to make 1 headband = Number of headbands Anne made / Total clothes used

= 6 ÷ 3/5

= 6 × 5/3

= (6 × 5) / 3

= 30/3

= 10 meters

Therefore, the amount of cloth needed to make 1 headband is 10 meters.

Learn more about unit value:

brainly.com/question/14286952

#SPJ1

4 0

2 years ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

3 years ago

( please help me) ( 11 points)

lara [203]

Answer:

500cm2

Step-by-step explanation:

2(9×20)+(7×20)

=360+140=500cm2

5 0

3 years ago