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3 years ago

Which geographical feature separates China and Mongolia?

Mathematics

2 answers:

Licemer1 [7]3 years ago

7 0

Answer:

Step-by-step explanation

The desert basins of the Gobi are bounded by the Altai Mountains and the grasslands and steppes of Mongolia on the north, by the Taklamakan Desert to the west, by the Hexi Corridor and Tibetan Plateau to the southwest and by the North China Plain to the southeast.

prohojiy [21]3 years ago

3 0

Answer:

The Gobi Desert

Step-by-step explanation:

The reason why its the Gobi Desert is because on the map it shows that the Gobi Desert is in between China and Mongolia.

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DochEvi [55]

Answer:

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3 years ago

C. If a < 0 and f(a) > 0, then the point (a, f(a))

professor190 [17]

The answer is: b - lies in quadrant II when graphed.

Explanation:

Any number less than zero is a negative number.

Example: -1

Any number greater than zero is a positive number.

Example: 1

(a, f(a))

(-1,1)

When these example points are graphed, they are placed in the second quadrant.

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2 years ago

1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help

Gnesinka [82]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Simplify :- 1 + - w² + 9w.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\large \sf1 + - w ^ { 2 } + 9 w$

Quadratic polynomial can be factored using the transformation $\sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ , where $\sf x_{1} and x_{2}$ are the solutions of the quadratic equation $\sf \: ax^{2}+bx+c=0$ .

$\large \sf-w^{2}+9w+1=0$

All equations of the form $\sf\:ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}$ . The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)} \\$

Square 9.

$\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)} \\$

Multiply -4 times -1.

$\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)} \\$

Add 81 to 4.

$\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)} \\$

Multiply 2 times -1.

$\large \sf \: w=\frac{-9±\sqrt{85}}{-2} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is plus. Add -9 to $\sf\sqrt{85}$ .

$\large \sf \: w=\frac{\sqrt{85}-9}{-2} \\$

Divide -9+ $\sf\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is minus. Subtract $\sf\sqrt{85}$ from -9.

$\large \sf \: w=\frac{-\sqrt{85}-9}{-2} \\$

Divide $\sf-9-\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}} \\$

Factor the original expression using $\sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ . Substitute $\sf\frac{9-\sqrt{85}}{2}$ for $\sf\:x_{1}$ and $\sf\frac{9+\sqrt{85}}{2}$ for $\sf\:x_{2}$ .

$\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}$

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

$\large \sf \: 1 + - w {}^{2} + 9w \\ =\large \boxed{\bf \: 1 - {w}^{2} + 9w}$