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fiasKO [112]
3 years ago
11

Which set of angle measures could be the measures of the interior angles of a triangle? please help me ​

Mathematics
1 answer:
marishachu [46]3 years ago
8 0

Answer:

B. 60-60-60

Step-by-step explanation:

The interior angles of a triangle must add up to 180 degrees. The only set of numbers adding up to 180 is choice B.

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What decimal is represented by this expanded form.? 4×10,000+5×1,000+7×100+2×1+8×1/100+6×1/1000
Nikolay [14]

The decimal would be 45702.086

6 0
3 years ago
Read 2 more answers
A line passes through tge points (1,-3) and (4,6) what is the zero of this line ?
Maru [420]

Answer:

<em>The zero of the line is (2,0)</em>

Step-by-step explanation:

The equation of a line passing through points (x1,y1) and (x2,y2) can be found as:

\displaystyle y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

We are given the points (1,-3) and (4,6). Substituting:

\displaystyle y+3=\frac{6+3}{4-1}(x-1)

Operating:

\displaystyle y+3=\frac{9}{3}(x-1)

y+3=3(x-1)

The zero of this line can be found by making y=0 and solving for x:

0+3=3(x-1)

3=3x-3

Adding 3:

6=3x

Solving:

x = 2

The zero of the line is (2,0)

3 0
3 years ago
I need help plzzzz, I’m really stupid
densk [106]

Answer:

d  sqrt(95)/7

Step-by-step explanation:

tan theta = opposite /adjacent

tan B = AC/BC

tan B = sqrt(95)/7

7 0
3 years ago
Find the greatest common factor of x^2y and xy^2
Dimas [21]
1. Find the H.C.F. of 4x2y3 and 6xy2z.
Solution:
The H.C.F. of numerical coefficients = The H.C.F. of 4 and 6.
Since, 4 = 2 × 2 = 22 and 6 = 2 × 3 = 21 × 31
Therefore, the H.C.F. of 4 and 6 is 2
6 0
4 years ago
show that thw roots of the equation (x-a)(x_b)=k^2 are always real if a,b and k are real. Please I really need help with this
VLD [36.1K]

Answer:

see explanation

Step-by-step explanation:

Check the value of the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real

• If b² - 4ac = 0 roots are real and equal

• If b² - 4ac < 0 then roots are not real

given (x - a)(x - b) = k² ( expand factors )

x² - bx - ax - k² = 0 ( in standard form )

x² + x(- a - b) - k² = 0

with a = 1, b = (- a - b), c = -k²

b² - 4ac = (- a - b)² + 4k²

For a, b, k ∈ R then (- a - b)² ≥ 0 and 4k² ≥ 0

Hence roots of the equation are always real for a, b, k ∈ R


           

8 0
3 years ago
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