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2 years ago

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Mathematics

1 answer:

lbvjy [14]2 years ago

6 0

Broo it’s blank for me....

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Please Help!!!!!!

navik [9.2K]

Answer:

6 team members

Step-by-step explanation:

We know that 2 team members can solve 13 problems in an hour. So, all you have to do is find what 195÷13 is, and multiply that by 2.

195÷13= 315

15*2= 30

And we have our answer!

I hope this helps! If it did, it would mean a lot to me if you could mark me brainliest :D

Have a nice day!

7 0

2 years ago

Read 2 more answers

Write the quadratic equation given the following points.<br> (1,-2) (2,-4) (3,-4)

pshichka [43]

Graph it and then tell it will come out

4 0

2 years ago

Write a polynomial function of least degree with integral coefficients that has the given zeros. 4, -6, 1

german

Answer: f(x) = x³ + x² - 26x + 24

<u>Step-by-step explanation:</u>

x = 4 x = -6 x = 1

x - 4 = 0 x + 6 = 0 x - 1 = 0

(x - 4) (x + 6) (x - 1) = 0

(x² + 2x - 24)(x - 1) = 0

x(x² + 2x - 24) -1(x² + 2x - 24) = 0

x³ + 2x² - 24x -x² - 2x + 24 = 0

x³ + x² - 26x + 24 = 0

7 0

2 years ago

How many congruent sides does a scalene triangle have?

Ainat [17]

The scalene triangle has 2 congruent sides

6 0

3 years ago

Read 2 more answers

There are three different hoses used to fill a pool: hose x, hose y, and hose z. Hose x can fill the pool in a days, hose y in b

WARRIOR [948]

Answer:

C) I and III only

Step-by-step explanation:

Let full pool is denoted by O

Days Hose x takes to fill pool O = a

Pool filled in one day x = O/a

Days Hose y takes to fill pool O = b

Pool filled in one day y = O/b

Days Hose z takes to fill pool O = c

Pool filled in one day z = O/c

It is given that

a>b>c

$a>b>c>d\\\implies x$

Days if if x+y+z fill the pool together = d

1 day if x+y+z fill the pool together $=O(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{O}{d}---(1)$

I) d < c

d are days when hose x, y, z are used together where as c are days when only z is used so number of days when three hoses are used together must be less than c when only z hose is used. So d < c

III) $\frac{c}{3}$

Using (1)

$\frac{bc+ac+ab}{abc}=\frac{1}{d}\\\\d=\frac{abc}{ab+bc+ca}\\\\As\quad(a>b>c)\\(ab+bc+ca)\frac{abc}{3ab}\\\\d>\frac{c}{3}$

Similarly

$\frac{bc+ac+ab}{abc}=\frac{1}{d}\\\\d=\frac{abc}{ab+bc+ca}\\\\As\quad a>b>c\\(ab+bc+ca)>3bc\\\\d=\frac{abc}{ab+bc+ca}$

So,

$\frac{c}{3}$

3 0

2 years ago