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2 years ago

Plz help me plz!!!!!(

Mathematics

1 answer:

Gre4nikov [31]2 years ago

5 0

A. 15 × 1 1/2= 15/1 × 3/2= 45/2= 2 2/5

b.15min +1extra minute=16min. × 15 students =240min so no not enough time

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Please help will give brainliest.

Olenka [21]

-1/7 is the correct answer for slope

4 0

2 years ago

Consider the expression below

brilliants [131]

This one is my real answer 5{10+3]=40 The other one you can use for extra this one 5{10+25]=35

4 0

2 years ago

Determine if the functions are even, odd or neither. f(x) = -5x4 - 2 and g(x) = x3 + 2x

Setler [38]

Try this option:
the rule: if f(-x)=f(x), then the function f(x) is even, if f(-x)=-f(x) then the function is odd.
1. f(x)=-5x⁴-2;
if to substitute x→(-x), then f(-x)=-5*(-x)⁴-2; ⇔ f(-x)=-5x⁴-2, in other words f(x)=f(-x), it means that this function is even.
2. f(x)=x³+2x.
if to substitute x→(-x), then f(-x)=(-x)³+2*(-x); ⇔ f(-x)=-(x³+2x), in other words f(-x)=-f(x), it means that this function is odd.

5 0

2 years ago

Given limit of f (x) = 4 as x approaches 0. what is limit of one-fourth left-bracket f (x) right-bracket superscript 4 baseline

Otrada [13]

The limit of the given function if $\lim_{x \to 0} f(x)=4$ is 64

<h3>Limit of a function</h3>

Given the following limit of a function expressed as;

$\lim_{x \to 0} f(x)=4$

We are to determine the value of the function

$\frac{1}{4} \lim_{x \to 0} [f(x)]^4$

This can also be expressed as

$\frac{1}{4} \lim_{x \to 0} [f(x)]^4\\ = \frac{1}{4}(4)^4 \\=1/4\times 256\\=64$

Hence the limit of the given function if $\lim_{x \to 0} f(x)=4$ is 64

Learn more on limit of a function here: brainly.com/question/23935467

#SPJ1

4 0

1 year ago

The functions f (theta) and g (theta) are sine functions, where f (0) equals g (0) equals 0.The amplitude of f (theta) is twice

777dan777 [17]

If period of $f(\theta)$ is one-half the period of $g(\theta)$ and
<span> $g(\theta)$ has a period of 2π, then $T_{g} =2T_{f}=2 \pi$ and $T_{f}= \pi$ .
</span>
To find the period of sine function $f(\theta)=asin(b\theta+c)$ we use the rule $T_{f}= \frac{2\pi}{b}$ .
<span /><span />
f is sine function where f (0)=0, then c=0; with period $\pi$ , then $f(\theta)=asin 2\theta$ , because $T_{f}= \frac{2 \pi }{2} = \pi$ .

To find a we consider the condition $f( \frac{ \pi }{4} )=4$ , from where $asin2* \frac{\pi}{4} =a*sin \frac{ \pi }{2} =a=4$ .

If the amplitude of $f(\theta)$ is twice the amplitude of $g(\theta)$ , then $g(\theta)$ has a product factor twice smaller than $f(\theta)$ and while period of $g(\theta)$ <span> </span> is 2π and g(0)=0, we can write $g(\theta)=2sin\theta$ .

8 0

2 years ago