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3 years ago

8

A lunch menu has 4 different sandwiches, 2 different soups, and 5 different drinks. How many different lunches consisting of a s

andwich, a soup, and a drink can you choose?

1 answer:

oksian1 [2.3K]3 years ago

5 0

To find out all of the possible combinations, just multiply the amount of each variable:

4*2*5 = 40

There are 40 different combinations consisting of a sandwich, soup, and drink.

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Make n the subject of m= n² + 3

Andrej [43]

The n subject of m = n² + 3 is $n=\sqrt{m-3}$

How is n made the subject of m = n² + 3 ?

m = n² + 3

n² = m- 3

$n=\sqrt{m-3}$

What does making n the subject mean?

To make n the subject of the formula or equation, we must rearrange the variables in the formula or equation to have a single variable n that is equivalent to the other variables in the formula.

The equation is then rearranged such that each term containing n is placed on the same side of the equation.

If n appears in several words, factorization might be necessary.

Just one n needs to be displayed at the conclusion to make it the subject.

The result of square rooting a number or variable as an inverse operation may be positive or negative.

To learn more about making n the subject , refer:

brainly.com/question/625155

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7 0

9 months ago

The lengths of the base and legs of an isosceles triangle are in the ratio of 3;2 the perimeter of triangles is 105 find the len

iVinArrow [24]

I guess you meam 3:2 instead of 3; 3
Anyhow,
The sides of isoceles triangle are leg, leg , base
each leg =2
while the base =4
To find the number you have to divide the ratio by total and multiply into 105
The base
3/7 × 105 = 45
The leg
2/7 × 105 = 30
Therefore, The length of the base = 45 cm

3 0

3 years ago

I need the answer. helpppppppppppp plzz

Liula [17]

Answer:lla te alldo pliss

Step-by-step explanation:

8 0

2 years ago

Enter the expression using exponents. 11 × 11 × 11 × 11 × 11 =

Alla [95]

Answer:

the answer should be $11^{5}$ or 11^5

Explanation:

Since there are five 11 in the equation, the answer should be $11^{5}$ or 11^5

You shouldn't write 161051 as the answer because it is NOT a exponent.

6 0

3 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

3 years ago