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2 years ago

10

Doug rents a kayak for 12 days the rental charges 18per day what is the best estimate for the total cost of the kayak rental

1 answer:

sashaice [31]2 years ago

8 0

Answer:

200?

Step-by-step explanation:

I did that by multiplying 20 by 10 because I rounded to those numbers.

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What is the value of expression?

kolezko [41]

$6^{-7}\cdot6^4=6^{-7+4}=6^{-3}$

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2 years ago

The square root of -12 × the square root of -9 (THIS IS AN ALGEBRA 2 QUESTION)

nlexa [21]

Answer:

$\sqrt{-12} = 2\sqrt{3} i\\\sqrt{-9}= 3i\\3i * 2\sqrt{3}i = -6\sqrt{3$

Step-by-step explanation:

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2 years ago

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

4 0

1 year ago

Jodi is going to start up a cookie baking business over the summer to make some extra money. It costs her $2.45 in supplies to m

Valentin [98]

Sfbxdfgbbffjxhndkxhbdbdidhdbdudbdjxudj

6 0

2 years ago

Find the center of the circle whose equation is x² + y² = 4.

Varvara68 [4.7K]

Answer:

The center of circle is $(0,0)$

Step-by-step explanation:

We need to find the center of the circle of the equation $x^{2}+y^{2}=4$

Since, the general equation of circle is $(x-h)^{2}+(y-k)^{2} = r^{2}$

Where (h,k) is center of circle and r is radius.

Re-write the circle equation is $x^{2}+y^{2}=4$ as,

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

Compare $x^{2}+y^{2}=2^{2}$ with $(x-h)^{2}+(y-k)^{2} = r^{2}$

so, $(x-0)^{2}+(y-0)^{2} = 2^{2}$

Hence, the center of circle is $(0,0)$

4 0

2 years ago

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