4[(6-1) +3 (5-2)] ????

Mathematics

1 answer:

kati45 [8]2 years ago

8 0

(6-1)+3(5-2)

5+3(5-2)

5+3*3

5+9

4*14

Answer=56

"Hope this helps"

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The hypotenuese of an isosceles right triangle is 16 inches. The midpoints of its sides are connected to form an inscribed trian

ser-zykov [4K]

Answer:

256/3 = 85 1/3 square inches

Step-by-step explanation:

The dimensions of the first inscribed triangle are 1/2 those of the original, so its area is (1/2)² = 1/4 of the original. The area of the original is ...

A = (1/2)bh = (1/2)(16/√2)(16/√2) = 64 . . . . square inches

The sum of an infinite series with first term 64 and common ratio 1/4 is ...

S = a1/(1 -r) . . . . . . for first term a1 and common ratio r

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3 years ago

Solve x2 - 8x - 9 = 0. Rewrite the equation so that it is of the form x2 + bx = c.

barxatty [35]

Answer:

x=9,-1 and $x^2+(-8)x=9$

Explanation:

we have been given with the quadratic equation $x^2-8x-9$

we compare the given quadratic equation with general quadratic equation

general quadratic is $ax^2+bx+c=0$

from given quadratic equation a=1,b= -8,c= -9

substituting these values in the formula for discriminant $D=b^{2}-4ac$

$D=(-8)^2-4(1)(-9)=100$

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Formula is $x=\frac{-b\pm\sqrt{D}}{2a}$

Now, substituting the values we will get

$x=\frac{-(-8)\pm\sqrt{100}} {2}= \frac{8\pm10}{2}=9,-1$

And rewritting the given equation by shifting 9 to right hand side of the given equation and taking minus inside the bracket so as to convert it in the form of

$x^2+bx=c$