1+0 free brainliest i really need help

2 answers:

8 0

<h2>✧･ﾟ: *✧･ﾟ:* Answer: *:･ﾟ✧*:･ﾟ✧ </h2><h3> </h3><h3>✅ 1 </h3><h3> </h3><h2>✧･ﾟ: *✧･ﾟ:* Explanation: *:･ﾟ✧*:･ﾟ✧ </h2><h3> </h3><h3>❗ 1 + 0 = 1 </h3><h3>the reason it isnt any other number is because 0 is in between negetave and positive numbers so 0 is nothing</h3><h3 /><h3>Example:</h3><h3>You have 1 cookie and you add nothing you still have 1 cookie! </h3><h3> </h3><h2>I WOULD APPRICIATE BRAINLIEST! </h2><h3> </h3><h3>~ ₕₒₚₑ ₜₕᵢₛ ₕₑₗₚₛ! :₎ ♡ </h3><h3> </h3><h3> </h3><h3>~ </h3>

IceJOKER [234]3 years ago

4 0

Answer:

.......

hope it helps !

have a great day!

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An object's height varies directly with the length of its shadow. A person who is 6 feet tall casts a 12-foot shadow. How long i

posledela

B.66 I am positively absolutely sure. Hope this helps!

8 0

3 years ago

Read 2 more answers

Simplify square root of ten times square root of eight.

Romashka-Z-Leto [24]

<h2>Explanation:</h2><h2></h2>

Here we need to simplify the square root of ten times square root of eight, so let's write this step by step:

Step 1. The square root of ten

$\sqrt{10}$

Step 2. The square root of eight

$\sqrt{8}$

Step 3. The square root of ten times square root of eight

$\sqrt{10}\times \sqrt{8} \\ \\ \\ \text{From radical property:} \\ \\ \sqrt{a}\times \sqrt{b}=\sqrt{ab} \\ \\ \\ So: \\ \\ \sqrt{10}\times \sqrt{8}=\sqrt{80} \\ \\ \\ But: \\ \\ \sqrt{80}=\sqrt{16\time5}=\sqrt{16}\sqrt{5}=4\sqrt{5}$

Finally, the correct option is:

<em>Four square root of five</em>

4 0

3 years ago

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Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess

Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2$