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2 years ago

Me. Gonzalez packs 45 boxes with limes. Each box holds 100 limes . How many limes can ms. Gonzalez pack into these boxes

Mathematics

2 answers:

GuDViN [60]2 years ago

4 0

Answer:

4,500

Step-by-step explanation:

45*100

Monica [59]2 years ago

3 0

Answer:

4,500

Step-by-step explanation:

To find the total number of limes, you need to multiply the group number (45 boxes) by the amount each can hold (100 limes).

45 x 100 = 4500 limes

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5x + 3y = -3 2X - 3y = 24

gregori [183]

Answer:

Y=-5/3x-1

y=2/3x-8

If you are adding then it is:

7x=21

x=3

Step-by-step explanation:

5x+3y=-3

3y=-5x-3

y=-5/3-1

2x-3y=24

-3y=-2x+24

y=2/3x-8

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To make 12 ounces of hot chocolate 3 tablespoons of cocoa are needed how many tablespoons of cocoa are needed to make 72 ounces

tiny-mole [99]

Answer:

you would need 18 tablespoons

Step-by-step explanation:

12--->72, so 3--->18

you are multiplying it by 6

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3 years ago

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a right trangle has one angle that measures 70 what is the measure of the other acute angle in the triangle

inna [77]

Answer:

Step-by-step explanation:

We know that the interior angle measures of a triangle are 180, and we are told that this is a right triangle.

Since we know the measure of two angles, we just have to do basic math:

180=90+70+x

solve for x (which equals 20) to get the measure of your acute angle

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3 years ago

What is m∠PQR? please help

leva [86]

Answer:

127

Step-by-step explanation:

The angles add up to 180° since they are on a straight line. Find 3x - 5.

3x - 5 + x + 1 = 180

4x - 4 = 180

4x = 180 - 4

= 176

x = 176 ÷ 4

x = 44

3x - 5 = 3(44) - 5

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3 years ago

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Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t

Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1. $f(x)=\frac{x^2+x-20}{x^2+4}$

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345$

3. $h(x)=\frac{3x-5}{x^2-5x+7}$

$x^2-5x+7=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

$\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43$

2. $g(x)=\frac{x-17}{x^2+75}$

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12$

4. $i(x)=\frac{x^2-9}{x-9}$

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is not continuous on the set of real numbers.

5. $j(x)=\frac{4x^2-7x-65}{x^2+10}$

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0$

6. $k(x)=\frac{x+1}{x^2+x+29}$

$x^2+x+29=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

$\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102$

7. $l(x)=\frac{5x-1}{x^2-9x+8}$

$x^2-9x+8=0$

$x^2-8x-x+8=0$

$x(x-8)-1(x-8)=0$

$(x-8)(x-1)=0$

$x=8,1$

The function is not defined for x=8 and x=1

Hence, function l is not defined for all real values.

8. $m(x)=\frac{x^2+5x-24}{x^2+11}$

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722$

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

6 0

3 years ago