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3 years ago

Write an algebraic expression for the words: "10 less than p".

Mathematics

2 answers:

poizon [28]3 years ago

7 0

"10 is less than P"
The algebraic expression for that would be:
10

(Hope this helps!)

olga55 [171]3 years ago

4 0

10 less than p:

10<p

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F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

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3 years ago

A quadrilateral is located on the coordinate plane with vertices at (2, 6), (2, −2), (4, −2), and (4, 6). Choose the vertices of

AURORKA [14]

Check the picture below.

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3 years ago

Help on 3 I do not get it

Rama09 [41]

The answer is nine hope this helps :)

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4 years ago

What is the slope of the line containing the points (1,-1) and (3,3)

marishachu [46]

Answer:

Step-by-step explanation:

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3 years ago

Tomika heard that the diagonals of a rhombus are perpendicular to each other. Help her test her conjecture. Graph quadrilateral

Stella [2.4K]

Answer:

a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus

b. The equation of the diagonal line AC is y = 5 - x

The equation of the diagonal line BD is y = 5 - x

c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other

Step-by-step explanation:

The vertices of the given quadrilateral are;

A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)

a. The length, l, of the sides of the given quadrilateral are given as follows;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

The length of side AB, with A = (1, 4) and B = (6, 6) gives;

$l_{AB} = \sqrt{\left (6-4 \right )^{2}+\left (6-1 \right )^{2}} = \sqrt{29}$

The length of side BC, with B = (6, 6) and C = (4, 1) gives;

$l_{BC} = \sqrt{\left (1-6 \right )^{2}+\left (4-6 \right )^{2}} = \sqrt{29}$

The length of side CD, with C = (4, 1) and D = (-1, -1) gives;

$l_{CD} = \sqrt{\left (-1-1 \right )^{2}+\left (-1-4 \right )^{2}} = \sqrt{29}$

The length of side DA, with D = (-1, -1) and A = (1,4) gives;

$l_{DA} = \sqrt{\left (4-(-1) \right )^{2}+\left (1-(-1) \right )^{2}} = \sqrt{29}$

Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus

b. The diagonals are AC and BD

The slope, m, of AC is given by the formula for the slope of a straight line as follows;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Therefore;

$Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1$

The equation of the diagonal AC in point and slope form is given as follows;

y - 4 = -1×(x - 1)

y = -x + 1 + 4

The equation of the diagonal AC is y = 5 - x

$Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1$

The equation of the diagonal BD in point and slope form is given as follows;

y - 6 = 1×(x - 6)

y = x - 6 + 6 = x

The equation of the diagonal BD is y = x

c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;

The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs

The point of intersection of the two diagonals is given as follows;

5 - x = x

∴ x = 5/2 = 2.5

y = x = 2.5

The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.

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3 years ago