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Arte-miy333 [17]
3 years ago
15

There are 20 horses in a field one quarter of the horses are brown how many brown horses are there

Mathematics
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer:

20 is total

1/4 of 20 = 5

5 horses are brown

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Math Questions :3 thank uu <3
coldgirl [10]
14.       1.5, 10    <- Answer
15.       5,1          <- Answer

Proof 14 

Solve the following system:
{2 x - y = -7 | (equation 1)
4 x - y = -4 | (equation 2)
Swap equation 1 with equation 2:
{4 x - y = -4 | (equation 1)
2 x - y = -7 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{4 x - y = -4 | (equation 1)
0 x - y/2 = -5 | (equation 2)
Multiply equation 2 by -2:
{4 x - y = -4 | (equation 1)
0 x+y = 10 | (equation 2)
Add equation 2 to equation 1:
{4 x+0 y = 6 | (equation 1)
0 x+y = 10 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 3/2 | (equation 1)
0 x+y = 10 | (equation 2)
Collect results:
Answer: {x = 1.5               
                y = 10

Proof 15. 

Solve the following system:
{5 x + 7 y = 32 | (equation 1)
8 x + 6 y = 46 | (equation 2)
Swap equation 1 with equation 2:
{8 x + 6 y = 46 | (equation 1)
5 x + 7 y = 32 | (equation 2)
Subtract 5/8 × (equation 1) from equation 2:{8 x + 6 y = 46 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Divide equation 1 by 2:
{4 x + 3 y = 23 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Multiply equation 2 by 4/13:
{4 x + 3 y = 23 | (equation 1)
0 x+y = 1 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 x+0 y = 20 | (equation 1)
0 x+y = 1 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 5 | (equation 1)
0 x+y = 1 | (equation 2)
Collect results:
Answer:  {x = 5                           y = 1
8 0
3 years ago
How do I solve this?
Gelneren [198K]
I hope this helps you! 

8 0
3 years ago
If the length of a rectangle is 3 and the width is 2, the length of the diagonal is a 13 b 5 c √5 d √13
trasher [3.6K]

Answer:

The length of the rectangle is 12cm and the area of the rectangle is 60cm2.

Explanation:

By definition, the angles of a rectangle are right. Therefore, drawing a diagonal creates two congruent right triangles. The diagonal of the rectangle is the hypotenuse of the right triangle. The sides of the rectangle are the legs of the right triangle. We can use the Pythagorean Theorem to find the unknown side of the right triangle, which is also the unknown length of the rectangle.

Recall that the Pythagorean Theorem states that the sun of the squares of the legs of a right triangle is equal to the square of the hypotenuse. a2+b2=c2

52+b2=132

25+b2=169

25−25+b2=169−25

b2=144

√b2=√144

b=±12

Since the length of the side is a measured distance, the negative root is not a reasonable result. So the length of the rectangle is 12 cm.

The area of a rectangle is given by multiplying the width by the length.

A=(5cm)(12cm)

A=60cm2

3 0
3 years ago
A culture started with 4,000 bacteria. After 2 hours it grew to 4400 bacteria. Predict how many bacteria will be present after 1
Art [367]
 <span>idk why they want the dratted "k" . it is so simple w/o it ! 
in the form y = ab^x, a = 4000, b = 4400/4000 = 1.1, x = t/2 
N = 4000*1.1^(13/2) = 7432 <------ 
----------------- 
now you won't be satisfied with simplicity, so 
4400 = 4000e^2k 
e^2k = 1.1 
k = ln 1.1/2 = .0477 to 4 dp 
N(13) = 4000e^(13*.0477) = 7436 <------ 

note that apart from a roundabout way, you get a less accurate ans !</span>
5 0
3 years ago
Read 2 more answers
Use the four-step definition of the derivative to find f'(x) if f(x) = −4x^3 −1.
guapka [62]

\stackrel{de finition \textit{ of a derivative as a limit}}{\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[-4(x+h)^3-1]~~ - ~~[-4x^3-1]}{h} \\\\\\ \cfrac{[-4(x^3+3x^2h+3xh^2+h^3)-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{[-4x^3-12x^2h-12xh^2-4h^3-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{-4x^3-12x^2h-12xh^2-4h^3-1+4x^3+1}{h}\implies \cfrac{-12x^2h-12xh^2-4h^3}{h}

\cfrac{h(-12x^2-12xh-4h^2)}{h}\implies -12x^2-12xh-4h^2 \\\\\\ \lim\limits_{h\to 0}~-12x^2-12xh-4h^2\implies \lim\limits_{h\to 0}~-12x^2-12x(0)-4(0)^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \lim\limits_{h\to 0}~-12x^2~\hfill

7 0
2 years ago
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