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3 years ago

Susan and Alejandro were in a race to finish their math work. Together they spent 30 minutes. Susan worked twice

Mathematics

2 answers:

EastWind [94]3 years ago

6 0

Answer:

susan spent 20 minutes alejandro spent 10

Step-by-step explanation:

S=A x2

30 divided by 3 is 10

susan gets 2/3 becasuse she worked 2 as fast alejandro gets 1/3

Rudik [331]3 years ago

3 0

Answer:

Susan worked for 10 mins

Step-by-step explanation:

because twice as fast 20 mins and 10 plus equal 30 I MIGHT BE WRONG

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Given cos B = 11/18 find angle B in degrees. Round your answer to the nearest hundredth.

Anastaziya [24]

$B = cos^{-1}(\frac{11}{18}) \approx 52.33^\circ$

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4 years ago

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West Jones bought markers for their teachers' dry erase boards. They bought 822 items in all. Each eraser cost $1.50 and each ma

Aleksandr-060686 [28]

Answer:

Number of erasers bought = 122

Number of markers bought = 700

Step-by-step explanation:

Given that:

Two items to be bought:

1. Eraser for the dry board

2. Maker

Total items bought = 822

Total money spent = $953

Cost of each eraser = $1.50

Cost of each marker = $1.10

Let the number of erasers bought = $x$

Let the number of markers bought = $y$

As per question statement:

$x+y=822 ..... (1)$

$1.50x+1.10y=953\\\Rightarrow 15x+11y=9530 ..... (2)$

Multiplying equation (1) with 11 and subtracting from equation (2):

$4x = 9530 - 822\times 11\\\Rightarrow 4x = 488\\\Rightarrow x=122$

By equation (1):

$\Rightarrow y =822-122\\\Rightarrow y = 700$

Therefore, the answer is:

Number of erasers bought = 122

Number of markers bought = 700

6 0

3 years ago

The Product of C and 8 is less than -18

Liono4ka [1.6K]

-18/8 = -2.25

therefore C is less than -2.25

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4 years ago

How to solve 3 ( -7 + e ) = 9

vampirchik [111]

Step-by-step explanation:

first we would do:

3 * -7 and e =

-21 + 3e

second

21+9= 30

so therefore e=10

sorry if this doesn't make sense

5 0

3 years ago

Which values of c will cause the quadratic equation –x2 + 3x + c = 0 to have no real number solutions? Check all that apply.

rodikova [14]

Answer:

-5

negative nine-halves

Step-by-step explanation:

we know that

In the quadratic equation $ax^{2} +bx+c=0$

If $b^{2}-4ac < 0$

then

The system has no real numbers solutions

we have

$-x^{2} +3x+c=0$

$a=-1,b=3$

substitute

$3^{2}-4(-1)c < 0$

$9+4c < 0$

$c < -\frac{9}{4}$

Verify each case

case 1) -5

For c=-5

substitute

$-5 < -\frac{9}{4}$

$-20 < 9$ ------> is true

therefore

The value of c=-5 will cause the quadratic equation to have no real number solutions

case 2) negative nine-halves

For c=-9/2

substitute

$-9/2 < -\frac{9}{4}$

$-36 < -18$ ------> is true

therefore

The value of c=-9/2 will cause the quadratic equation to have no real number solutions

case 3) negative one-quarter

For c=-1/4

substitute

$-1/4 < -\frac{9}{4}$

$-4 < -36$ ------> is not true

therefore

The value of c=-1/4 will not cause the quadratic equation to have no real number solutions

case 4) 1

For c=1

substitute

$1 < -\frac{9}{4}$ ------> is not true

therefore

The value of c=1 will not cause the quadratic equation to have no real number solutions

case 5) 9 Over 4

For c=9/4

substitute

$9/4< -\frac{9}{4}$ ------> is not true

therefore

The value of c=9/4 will not cause the quadratic equation to have no real number solutions

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3 years ago

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