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3 years ago

Sharon is planning a holiday for 4 people for 7 days.

Mathematics

2 answers:

OleMash [197]3 years ago

7 0

Answer: 3600

Step-by-step explanation:

(Travel + Hotel + Spending money ) x 4 people=

130 + 60 x 7 (days) + 350 = 900 x 4 = 3600

Eduardwww [97]3 years ago

6 0

1 person = 130 + (60x7) + 350 = 900
4 people = 4 x 900 = 3600
I think this is right !

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The area of the rectangle as a single expression

Ne4ueva [31]

Answer: $A=15x^2+6x-48$

Step-by-step explanation:

The area of a rectangle can be calculated with this formula:

$A=l*w$

Where "l" is the length and "w" is the width.

You can identify from the figure that the length and the width of this rectangle are:

$length=3(x+2)\\width=5x-8$

Then, you need to substitute this lenght and this width into $A=l*w$ :

$A=3(x+2)(5x-8)$

Now, apply Distributive property:

$A=(3x+6)(5x-8)\\\\A=15x^2-24x+30x-48$

Finally, you need to add the like terms. Then, you get:

$A=15x^2+6x-48$

5 0

3 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

5 0

2 years ago

Read 2 more answers

The product of two whole numbers is 608 and their sum is 51 .what are the two numbers ?

amid [387]

Answer:

The product of 19 and 32 is 608 and their sum adds up to 51.

The two numbers are: 19 and 32

3 0

2 years ago

20% percent tip on 8 dollar

sesenic [268]

Answer:

9.60 dollars tip is $1.60

Step-by-step explanation:

8 0

2 years ago

Find the slope of a line perpendicular to y = 3x + 7

Vikki [24]

Answer:

-1/3

Step-by-step explanation:

The slope of the line has the form y = mx + b where m is the slope. Here the slope is m = 3. A line perpendicular to it will have a negative reciprocal slope. The negative reciprocal of 3 is -1/3.

4 0

3 years ago