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3 years ago

Help, need index form

Mathematics

1 answer:

ivolga24 [154]3 years ago

7 0

Refer to the attachment for complete calculation.

3²× 11^-9

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A man jogs to the shop and back in half an hour on the way he jogs at 4m/s and on the way back he jogs at 5m/s ignoring time spe

pochemuha

Ratio 4:5
There are 9 total.
30 mins = 1800 secs.
1800 divided by 9 = 200
200 secs = 3mins 20secs
200 is one "part" of the ratio, so multiply it by 4, then 5
800:1000
This is the important one
V
13mins, 20secs: 16mins 40secs
Distance is speed x time,
16mins x 4 = 64
40 secs = 2/3
x 4 = 1.6666
Distance to shop - 65.6m

5 0

3 years ago

Which is a quadratic function having a leading coefficient of 3 and a constant term of –12? f(x) = –12x2 3x 1 f(x) = 3x2 11x – 1

eimsori [14]

The correct option is b.

<h2>Quadratic Equation</h2>

A quadratic equation is an equation that can be written in the form of

ax²+bx+c.

Where a is the leading coefficient, and

c is the constant.

<h2>Leading coefficient </h2>

The leading coefficient of an equation is a coefficient of the term having the highest power in the equation.

Given options:

a.) $f(x) = -12x^2+3x+1 \\$

b.) $f(x) = 3x^2+11x-12 \\$

c.) $f(x) = 12x^2+3x+3\\$

d.) $f(x) = 3x -12.$

<h3 /><h3>Solution</h3>

As it is given that the leading coefficient of the equation is 3. therefore, the value of the 'a' in the general quadratic equation is 3,

3x²+bx+c,

Also, the value of the constant is -12, therefore, the value of c is -12,

3x²+bx-12

From all the given options the only feasible option is b only.

Hence, the correct option is b.

Learn more about Quadratic Equation:

brainly.com/question/2263981

8 0

2 years ago

A hog producer is feeding soybean meal and corn to his pigs. He needs to know how many pounds he needs to feed of each in order

Mazyrski [523]

Answer:

Feed 1, Soybean meal required = $\dfrac{2000}{3} \text{ lbs.}$

Feed 2, Corn meal required = $\dfrac{4000}{3} \text{ lbs.}$

Step-by-step explanation:

Total feed is <em>1 ton</em> i.e. <em>2000 lbs</em>.

Let <em>x</em> be the amount of Feed 1 required.

Feed 1 has $45\%$ of protein.

$\text{Protein in Feed 1 = }x \times \dfrac{45}{100} ..... (1)$

Then, amount of Feed 2 required = $(2000 - x) \text{ lbs.}$

Feed 2 has $15\%$ of protein.

$\text{Protein in Feed 2 = }(2000 - x) \times \dfrac{15}{100} ..... (2)$

As per question, total protein required is $25\%$ of 2000 lbs .

Adding (1) and (2) and putting it equal to total protein required.

$\Rightarrow x \times \dfrac{45}{100} + (2000 - x) \times \dfrac{15}{100} = 2000 \times \dfrac{25}{100}\\\Rightarrow 45x + 30000 - 15x = 50000\\\Rightarrow 30x = 20000\\\Rightarrow x = \dfrac{2000}{3}$