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3 years ago

Is the quadrilateral a parallelogram? No Yes Not enough info

Mathematics

2 answers:

taurus [48]3 years ago

8 0

Yes, it can be a parallelogram. This is because both sides are parallel.

Aleonysh [2.5K]3 years ago

6 0

Yes because both sides are parallel

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3 years ago

Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15.

Naya [18.7K]

It's a geometric sequence.

$4,-12,36,... \\ \\
a_1=4 \\
r=\frac{a_2}{a_1}=\frac{-12}{4}=-3 \\ \\
a_n=a_1 \times r^{n-1} \\
a_n=4 \times (-3)^{n-1} \\
a_n=4 \times (-3)^{-1} \times (-3)^n \\
a_n=4 \times (-\frac{1}{3}) \times (-3)^n \\
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It's the sum for term 4 through term 15.

$\boxed{ \sum\limits_{n=4}^{15} (\frac{4}{3}(-3)^n)}$

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3 years ago

Subtract (2x3 + 9x − 8) − (4x2 − 15x + 7).

STatiana [176]

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3 years ago

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3 years ago

The point P(7, −2) lies on the curve y = 2/(6 − x). (a) If Q is the point (x, 2/(6 − x)), use your calculator to find the slope

NARA [144]

Answer:

a) (i) $m = 2.22$ , (ii) $m = 2$ , (iii) $m = 2$ , (iv) $m = 2$ , (v) $m = 1.82$ , (vi) $m = 2$ , (vii) $m = 2$ , (viii) $m = 2$ ; b) $m \approx 2$ ; c) The equation of the tangent line to curve at P (7, -2) is $y = 2\cdot x + 12$ .

Step-by-step explanation:

a) The slope of the secant line PQ is represented by the following definition of slope:

$m = \frac{\Delta y}{\Delta x} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}$

(i) $x_{Q} = 6.9$ :

$y_{Q} =\frac{2}{6-6.9}$

$y_{Q} = -2.222$

$m = \frac{-2.222 + 2}{6.9-7}$

$m = 2.22$

(ii) $x_{Q} = 6.99$

$y_{Q} =\frac{2}{6-6.99}$

$y_{Q} = -2.020$

$m = \frac{-2.020 + 2}{6.99-7}$

$m = 2$

(iii) $x_{Q} = 6.999$

$y_{Q} =\frac{2}{6-6.999}$

$y_{Q} = -2.002$

$m = \frac{-2.002 + 2}{6.999-7}$

$m = 2$

(iv) $x_{Q} = 6.9999$

$y_{Q} =\frac{2}{6-6.9999}$

$y_{Q} = -2.0002$

$m = \frac{-2.0002 + 2}{6.9999-7}$

$m = 2$

(v) $x_{Q} = 7.1$

$y_{Q} =\frac{2}{6-7.1}$

$y_{Q} = -1.818$

$m = \frac{-1.818 + 2}{7.1-7}$

$m = 1.82$

(vi) $x_{Q} = 7.01$

$y_{Q} =\frac{2}{6-7.01}$

$y_{Q} = -1.980$

$m = \frac{-1.980 + 2}{7.01-7}$

$m = 2$

(vii) $x_{Q} = 7.001$

$y_{Q} =\frac{2}{6-7.001}$

$y_{Q} = -1.998$

$m = \frac{-1.998 + 2}{7.001-7}$

$m = 2$

(viii) $x_{Q} = 7.0001$

$y_{Q} =\frac{2}{6-7.0001}$

$y_{Q} = -1.9998$

$m = \frac{-1.9998 + 2}{7.0001-7}$

$m = 2$

b) The slope at P (7,-2) can be estimated by using the following average:

$m \approx \frac{f(6.9999)+f(7.0001)}{2}$

$m \approx \frac{2+2}{2}$

$m \approx 2$

The slope of the tangent line to the curve at P(7, -2) is 2.

c) The equation of the tangent line is a first-order polynomial with the following characteristics:

$y = m\cdot x + b$

Where:

$x$ - Independent variable.

$y$ - Depedent variable.

$m$ - Slope.

$b$ - x-Intercept.

The slope was found in point (b) (m = 2). Besides, the point of tangency (7,-2) is known and value of x-Intercept can be obtained after clearing the respective variable:

$-2 = 2 \cdot 7 + b$

$b = -2 + 14$

$b = 12$

The equation of the tangent line to curve at P (7, -2) is $y = 2\cdot x + 12$ .

7 0

3 years ago