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3 years ago

2. Solve the following equations for 0° Sx < 180°.

Mathematics

1 answer:

diamong [38]3 years ago

5 0

Answer:

See below in bold.

Step-by-step explanation:

(1) cosec ( x + 10) = 3

Now cosec x = 1/sin x so

1 / sin (x + 10) = 3

3 sin (x + 10) = 1

sin (x + 10) = 1/3

x + 10 = 19.47 , 160.53 degrees

x = 9.47, 150.53 degrees.

(ii) cot (x - 30) =0.45

cot (x - 30)= 1 /tan (x- 30) so we have

tan (x - 30) = 1 / 0.45 = 2.2222

x - 30 = 65.77 degrees

x = 95.77 degrees.

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Which representation of a transformation on a coordinate grid does not preserve congruence?

Alex17521 [72]

Answer:

Step-by-step explanation:

Consider all options:

A. Transformation with the rule

$(x,y)\rightarrow (x,-y)$

is a reflection across the x-axis.

Reflection across the x-axis preserves the congruence.

B. Transformation with the rule

$(x,y)\rightarrow \left(\dfrac{7}{8}x,\dfrac{7}{8}y\right)$

is a dilation with a scale factor of $\frac{7}{8}$ over the origin.

Dilation does not preserve the congruence as you get smaller figure.

C. Transformation with the rule

$(x,y)\rightarrow (x+6,y-4)$

is a translation 6 units to the right and 4 units down.

Translation 6 units to the right and 4 units down preserves the congruence.

D. Transformation with the rule

$(x,y)\rightarrow (y,-x)$

is a clockwise rotation by $90^{\circ}$ angle over the origin.

Clockwise rotation by $90^{\circ}$ angle over the origin preserves the congruence.

3 0

3 years ago

Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the

Rama09 [41]

Question:

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s 2s 3s 4s

Answer:

It takes 2 seconds for object to hit the ground

Solution:

The given equation is:

$h(t) = -16t^2 + v_0t+h_0$

Initial velocity = 27 feet/sec

$h_0 = 10\ feet$

Therefore,

$h(t) = -16t^2 +27t+10$

At the point the object hits the ground, h(t) = 0

$-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0$

Solve by quadratic formula,

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125$