Answer:
-1
Step-by-step explanation:
We are given the expression:
-5x + 6y - 7y + 4x
With condition x = -3, y = 4
Therefore, substitute x = -3 and y = 4.
-5(-3) + 6(4) - 7(4) + 4(-3)
Recall every important fundamental math such as multiplying with negative.
15 + 24 - 28 - 12
Evaluate:
39 - 40
-1
Alternative solution is to combine like terms first before substitution.
-x - y
Thus:
-(-3) - 4
3 - 4
-1
Step-by-step explanation:
LHS:
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{\sqrt{8}+\sqrt{9}}1+21+2+31+3+41+……………+8+91
Rationalizing the denominator, we get
\Rightarrow\left(\frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}+\sqrt{4}} \times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}\right)+\cdots \ldots+\left(\frac{1}{\sqrt{8}+\sqrt{9}} \times \frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}\right)⇒(1+21×1−21−2)+(2+31×2−32−3)+(3+41×3−43−4)+⋯…+(8+91×8−98−9)
We know that,
\left(a^{2}-b^{2}\right)=(a+b)(a-b)(a2−b2)=(a+b)(a−b)
Now, on substituting the formula, we get,
=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\cdots \ldots \cdot \frac{(\sqrt{8}-\sqrt{9})}{8-9}=1−21−2+2−32−3+3−43−4+⋯…⋅8−9(8−9)
\Rightarrow \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{8}+\sqrt{9}}=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{9}-\sqrt{8})⇒1+21+
A compass because youre just making two arcs
420:540 or 70:90 or 35:45 or 7:9
Choose either one if I'm wrong I'm sorry
