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3 years ago

Dena has 10 flowers. there are two kinds

Mathematics

1 answer:

Simora [160]3 years ago

8 0

5 is what I know ..........................

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random question if where humans dose that mean were a creature to bugs so confuse someone help the will get brainlist

kondaur [170]

In theory, any living organism is a creature because it needs the same basic things to live. so yes, I think us humans are a creature to bugs.

6 0

3 years ago

This week, 951 people went to the beach. Last week, 1,173 people went to the beach. The number of people who went to the beach

Ugo [173]

It fell by 19%
Use 1173 as your total and then find the percent from 951 and subtract from 100 and that’s the percent that fell

4 0

3 years ago

The points (6, -23) a

ZanzabumX [31]

Answer:

$r=-3$

Step-by-step explanation:

The line that passes through the two points (6, -23) and (11, r) has a slope of 4.

We want to determine the missing value of r.

We can use the slope formula:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Let (6, -23) be (x₁, y₁) and let (11, r) be (x₂, y₂). The final answer should be 4. Therefore:

$\displaystyle \frac{r-(-23)}{11-6}=4$

Simplify and evaluate:

$\displaystyle \frac{r+23}{5}=4$

Multiply both sides by 5:

$r+23=20$

Therefore:

$r=-3$

4 0

3 years ago

The equation of a circle whose center is at (1,2) and radius is 5 is (x + 1)² + (y + 2)² = 5 (x - 1)² + (y - 2)² = 25 (x - 1)² -

Oksana_A [137]

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{1}{ h},\stackrel{2}{ k})\qquad \qquad 
radius=\stackrel{5}{ r}
\\\\\\\
(x-1)^2+(y-2)^2=5^2\implies (x-1)^2+(y-2)^2=25$

7 0

3 years ago

Read 2 more answers

Alenkasestr [34]

We start with the expression at the left of the equation.

We can combine the terms as:

$\begin{gathered} \frac{2+\sqrt[]{3}}{\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}}}-\frac{2-\sqrt[]{3}}{\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}}} \\ \frac{2+\sqrt[]{3}}{\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}}}\cdot\frac{(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})}-\frac{2-\sqrt[]{3}}{\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}}}\cdot\frac{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})} \end{gathered}$

We can now apply the distributive property for the both the numerator and denominator. We can see also that the denominator is the expansion of the difference of squares:

$\begin{gathered} \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2})^2-(\sqrt[]{2-\sqrt[]{3}}))^2} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})+(\sqrt[]{3}-2)\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{2^{}-(2-\sqrt[]{3})^{}} \\ \frac{\sqrt[]{2}\cdot(2+\sqrt[]{3})-\sqrt[]{2-\sqrt[]{3}}\cdot(2+\sqrt[]{3})+\sqrt[]{2}\cdot(\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}\cdot(\sqrt[]{3}-2)}{2-2+\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2+\sqrt[]{3}+\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}(-2-\sqrt[]{3}+\sqrt[]{3}-2)}{\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2\sqrt[]{3})+\sqrt[]{2-\sqrt[]{3}}(-4)}{\sqrt[]{3}} \\ 2\sqrt[]{2}-4\frac{\sqrt[]{2-\sqrt[]{3}}}{\sqrt[]{3}} \end{gathered}$

We then can continue rearranging this as:

7 0

1 year ago