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Whitepunk [10]
3 years ago
8

Perform the operation and reduce the answer fully. 11/20 - 7/12​

Mathematics
1 answer:
zvonat [6]3 years ago
6 0
- 1/ 30 !!
Hope I helped ! ❤️❤️
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If 0 and 3 is equal to y = 30<br> what is 3 and 5.5 equal to
Delvig [45]

Answer:

8.5

Step-by-step explanation:

3 0
3 years ago
What is the fraction ofr 0.0625?
Sunny_sXe [5.5K]
The fraction of 0.0625 is 1/16
4 0
3 years ago
Read 2 more answers
If one of the purple-flowered progeny plants is selected at random and self-fertilized, what is the probability it will breed tr
777dan777 [17]
This is in reality a biology question given a lot of biology related information is not given in the problem.
However, I will provide the answer by making biology-related assumptions.
1. The genotype of the purple-flowered progeny is Pp
2. The purple-flower phenotype is recessive.

If self fertilized, both "parents" have genotype Pp
The punnett square is as follows:

         P      p
P     PP    Pp
p     Pp     pp

Since purple-flowers are recessive, only PP will come out purple, hence the probability is 1/4.
4 0
3 years ago
Solve by completing the square <br> please solve all questions 1-9 for 50 points + brainly
Alenkinab [10]

\Large{\underline{\underline{{\mathfrak{{\bigstar}\:Answer}}}}}\\\\

\bf{1.\:\:x^{2}+4x+1=0}

\longrightarrow\:\:\sf{x=\dfrac{-4{\underline{+}}\sqrt{4^{2}-4\times1\times1}}{2\times1}}

\longrightarrow\:\:\sf{x=\dfrac{-4{\underline{+}}\sqrt{16-4}}{2}}

\longrightarrow\:\:\sf{x=\dfrac{-4{\underline{+}}\sqrt{12}}{2}}

\longrightarrow\:\:\sf{x=\dfrac{4{\underline{+}}2\sqrt{3}}{2}}

\longrightarrow\:\:\sf{x_{1}=-2-\sqrt{3}\:,\:x_{2}= -2+\sqrt{3}}\\\\\\

\bf{2.\:\:x^{2}-9x+14=0}

\longrightarrow\:\:\sf{x^{2}-2x-7x+14=0}

\longrightarrow\:\:\sf{x\times (x-2)-7(x-2)=0}

\longrightarrow\:\:\sf{x_{1}=2\:,\:x_{2}=7}\\\\\\

\bf{3.\:\:x^{2}+8x+2=22}

\longrightarrow\:\:\sf{x^{2}+8x+2-22=0}

\longrightarrow\:\:\sf{x(x+10)-2(x+10)=0}

\longrightarrow\:\:\sf{(x+10)(x-2)=0}

\longrightarrow\:\:\sf{x_{1}=-10\:,\:x_{2}=2}\\\\\\

\bf{4.\:\:x^{2}+8x+7=0}

\longrightarrow\:\:\sf{x^{2}+7x+x+7=0}

\longrightarrow\:\:\sf{x(x+7)+x+7=0}

\longrightarrow\:\:\sf{(x+7)(x+1)=0}

\longrightarrow\:\:\sf{x_{1}=-7\:,\:x_{2}=1}\\\\\\

\bf{5.\:\:x^{2}-10x+25=9}

\longrightarrow\:\:\sf{(x-5)^{2}=9}

\longrightarrow\:\:\sf{x-5={\underline{+}}3}

\longrightarrow\:\:\sf{x-5=-3 \: ; \:x-5=3}

\longrightarrow\:\:\sf{x_{1}=2\:,\:x_{2}=8}\\\\\\

\bf{6.\:\:x^{2}-10x+16=0}

\longrightarrow\:\:\sf{x^{2}-2x-8x+16=0}

\longrightarrow\:\:\sf{(x-2)(x-8)=0}

\longrightarrow\:\:\sf{x-2=0\:;\:x-8=0}

\longrightarrow\:\:\sf{x_{1}=2\:,\:x_{2}=8}\\\\\\

\bf{7.\:\:2x^{2}+7x-4=0}

\longrightarrow\:\:\sf{2x(x+4)-(x-4)=0}

\longrightarrow\:\:\sf{(x+4)((2x-1)=0}

\longrightarrow\:\:\sf{x+4=0\:;\:2x-1=0}

\longrightarrow\:\:\sf{x_{1}=-4\:,\:x_{2}=\dfrac{1}{2}}\\\\\\

\bf{8.\:\:x^{2}-2x+3=0}

\longrightarrow\:\:\sf{x=\dfrac{-(-2){\underline{+}}\sqrt{(-2)^{2}-4\times 1\times 3}}{2\times1}}

\longrightarrow\:\:\sf{\dfrac{2{\underline{+}}\sqrt{4-12}}{2}}

\longrightarrow\:\:\sf{\dfrac{2{\underline{+}}\sqrt{-8}}{2}}

\longrightarrow\:\:\sf{x \notin {\mathbb{R}}}\\\\\\

\bf{9.\:\:3x^{2}+8x+5=0}

\longrightarrow\:\:\sf{x(3x+5)+3x+5=0}

\longrightarrow\:\:\sf{(3x+5)(x+1)=0}

\longrightarrow\:\:\sf{3x+5=0\:;\:x+1=0}

\longrightarrow\:\:\sf{x_{1}=-\dfrac{5}{3}\:,\:x_{2}=-1}\\\\\\

\bf{10.\:\:{\mathbb{\red{1}}}\:\:}:')

5 0
2 years ago
What is the solution for the square root of x&lt;9
nydimaria [60]

The square root only accepts non-negative inputs, so the solution will surely be a non-negative value: x \geq 0

Also, the square root of 81 is exactly 9: \sqrt{81} = 9.

Finally, we have to observe that the square root grows with its inputs:

x > y > 0 \implies \sqrt{x} > \sqrt{y}

So, you want the square root of a number to be less than 9, and you know that the square root of 81 is exactly 9, and that the square root gets smaller as the inputs get smaller, and larger as the inputs get larger.

So, the answer is 0 \leq x < 81

If you want to be more rigorous, you can observe that since both sides of the inequality are positive you can square them:

\sqrt{x}

But you can't accept this solution, because you have to remember that x has to be positive, so you have to correct it into

0 \leq x < 81

Just as before

3 0
3 years ago
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