All categories

3 years ago

Pls help I will give Brainlyest

Mathematics

1 answer:

baherus [9]3 years ago

8 0

Answer:

FALSE

Step-by-step explanation:

If p could equal 10, 20, 30 or 40 lets put each of them in the expression 7.4p<=74

1. 7.4 times 10 equals 74 so that work✅

2. 7.4 times 20 equals 148 so that does not work X

3. 7.4 times 30 equals 222 so that does not work X

4. 7.4 times 40 equals 296 so that does not work X

You might be interested in

Three students share 28 pencils equally.how many pencils does each person get?

Nonamiya [84]

Answer:

Step-by-step explanation:

28 divided by 7=4

3 0

3 years ago

975 toffees are to be distributed among 23 children. Estimate the numbers of toffees that each child will get in nearest tens.

Aneli [31]

Answer:

Step-by-step explanation:

Total number of toffees = 975

Total number of children = 23

Estimate the numbers of toffees that each child will get in nearest tens.

Number of toffees each child get = Total number of toffees / Total number of children

= 975 / 23

= 42.391304347826

Approximately,

Number of toffees each child get = 40 to the nearest tens

3 0

3 years ago

Which expression can be used to find 7

mariarad [96]

Answer:

7(-6) +7

Step-by-step explanation:

hope it helps

have a good day

5 0

2 years ago

If P(A)=0.3, then the probability of the complement of A is ?

KonstantinChe [14]

C is the answer for this question

7 0

3 years ago

Determine the horizontal vertical and slant asymptote y=x^2+2x-3/x-7

lilavasa [31]

Answer:

<h2>A.Vertical:x=7</h2><h2>Slant:y=x+9</h2>

Step-by-step explanation:

$f(x)=\dfrac{x^2+2x-3}{x-7}\\\\vertical\ asymptote:\\\\x-7=0\qquad\text{add 7 to both sides}\\\\\boxed{x=7}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{1-\frac{7}{x}}=\pm\infty\\\\\boxed{not\ exist}$

$slant\ asymptote:\\\\y=ax+b\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{f(x)}{x}\\\\b=\lim\limits_{x\to\pm\infty}(f(x)-ax)\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x^2+2x-3}{x-7}}{x}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x(x-7)}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x^2-7x}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x^2\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{1+\frac{2}{x}-\frac{3}{x^2}}{1-\frac{7}{x}}=\dfrac{1}{1}=1$

$b=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-1x\right)=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x(x-7)}{x-7}\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x^2-7x}{x-7}\right)=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-(x^2-7x)}{x-7}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-x^2+7x}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{9x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(9-\frac{3}{x}\right)}{x\left(1-\frac{7}{x}\right)}$

$=\lim\limits_{x\to\pm\infty}\dfrac{9-\frac{3}{x}}{1-\frac{7}{x}}=\dfrac{9}{1}=9\\\\\boxed{y=1x+9}$

8 0

3 years ago