Which of the diagrams below represents the statement "If it is not a gorilla, then it is not an ape"?

Mathematics

2 answers:

DaniilM [7]3 years ago

8 0

Comment is correct, I mis read. it is figure B since not gorilla is inside not ape

tekilochka [14]3 years ago

5 0

Answer: Figure B, just did on apex

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Please help with this question!!

xxMikexx [17]

$\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2\\\\l\perp k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\\text{We have the points J(-24, -4) and K(-4, 6)}.\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{substitute:}\\\\m_1=\dfrac{6-(-4)}{-4-(-24)}=\dfrac{10}{20}=\dfrac{1}{2}\\\\\text{therefore}\ m_2=-\dfrac{1}{\frac{1}{2}}=-2\\\\\text{The formula of a midpoint:}\\\\\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)\\\\\text{substitute the coordinates of the points J and K:}$

$x=\dfrac{-24+(-4)}{2}=\dfrac{-28}{2}=-14\\\\y=\dfrac{-4+6}{2}=\dfrac{2}{2}=1\\\\\text{midpoint}\ (-14,\ 1)\\\\\text{The point-slope form:}\\\\y-y_1=m(x-x_1)\\\\\text{substitute}\ m=-2,\ x_1=-14\ \text{and}\ y_1=1:\\\\y-1=-2(x-(-14))\\\\\boxed{y-1=-2(x+14)}$

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3 years ago

The fourth and fifth grade classes at James Madison Elementary School are going on a field trip. There are 87 students in fourth

Anna11 [10]

Answer:

about 7

Step-by-step explanation:

6 0

3 years ago

Read the following two statements. Then, if possible, use the Law of Detachment to draw a conclusion.

jeyben [28]

<span>The hypothesis is : If Robbie wants to save money to buy a car. Conclusion : he must get a part-time job.

Robbie already started the new job yesteday. So, that makes the hypothesis true.

So, the conclusion would be : Robbie will save money to buy a car/ Robbie will buy a car.</span>

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3 years ago

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I need help. this is the problem x^2/3+10=7x^1/3

Arlecino [84]

$\it x^{\frac{2}{3}}+10=7x^{\frac{1}{3}} \Leftrightarrow (x^{\frac{1}{3}})^2-7x^{\frac{1}
 {3}} +10=0 
\\\;\\
We\ note\ x^{\frac{1}{3}} =t \ \ and \ the \ equation \ will \ be:
\\\;\\
t^2-7t+10 = 0 \Leftrightarrow t^2 -2t-5t+10=0 \Leftrightarrow t(t-2) -5(t-2)=0$

$\it \Leftrightarrow (t-2)(t-5)=0 
\\\;\\
t-2=0 \Rightarrow t=2 \Rightarrow x^{\frac{1}{3}}=2 \Rightarrow (x^{\frac{1}{3}})^3 =2^3 \Rightarrow x = 8
\\\;\\
t-5=0 \Rightarrow t=5 \Rightarrow x^{\frac{1}{3}}= 5 \Rightarrow (x^{\frac{1}{3}})^3 = 5^3 \Rightarrow x = 125$

S = {8; 125}

7 0

3 years ago

Please and thank you

Alina [70]

The answer is -1 and 0

6 0

3 years ago

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