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3 years ago

PLEASE HELP. BRAINLEST IF CORRECT

Mathematics

2 answers:

atroni [7]3 years ago

8 0

Answer:

yes is the answer

hope it helps you

Mademuasel [1]3 years ago

8 0

Answer:

4/9

Step-by-step explanation:

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Please answer this correctly

dexar [7]

Answer:

It’s a 1/2 chance it’s heads.

Step-by-step explanation:

Because there’s two sides

4 0

4 years ago

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Min is twice as tall as her brother . If min is 5feet 3 inches tall how tall is her brother

liubo4ka [24]

5’3 is 63 inches since her brother is 1/3 larger so 63/3= 21 inches

3 0

4 years ago

Helppppp! Angles PTQ and STR are vertical angles and congruent. Circle T is shown. Line segments T P, T Q, T R, and T S are radi

cestrela7 [59]

Answer:

$\overline{QP} = \overline{SR}$

Step-by-step explanation:

Given: ∠PTQ ≅ ∠STR and TP, TQ, TR and TS are radii of the circle.

If two angle are equal, then they are equal in size and their measurements are equal.

IF ∠PTQ ≅ ∠STR then they are also equal in their measurements.

Now, In ΔPTQ and ΔSTR

∠PTQ ≅ ∠STR (vertically opposite angle)

$\overline{TQ} \cong \overline{ST}$ (Radius of the circle are equal and congruent in length)

$\overline{PT} \cong \overline{TR}$ (Radius of the circle)

∴ ΔPTQ ≅ ΔSTR (SAS postulate)

∴ $\overline{QP}\cong \overline{SR}$ (corresponding sides of congruent triangle are equal and congruent).

6 0

3 years ago

Read 2 more answers

Use the table below.

Alexandra [31]

Answer:

50 minutes.

Step-by-step explanation:

7*50 = 350

8 0

3 years ago

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Consider the transformation T: x = \frac{56}{65}u - \frac{33}{65}v, \ \ y = \frac{33}{65}u + \frac{56}{65}v

irina1246 [14]

Answer:

Step-by-step explanation:

$T: x = \frac{56}{65}u - \frac{33}{65}v, \ \ y = \frac{33}{65}u + \frac{56}{65}v$

$\frac{d(x,y)}{d(u,v)} =\left|\begin{array}{ccc}x_u&x_v\\y_u&y_v\end{array}\right|$

$=(\frac{56}{65} )^2+(\frac{33}{65} )^2\\\\=\frac{(56)^2+(33)^2}{(65)^2} \\\\=\frac{4225}{4225} \\\\=1$

B )

$S:-65 \leq u \leq 65, -65 \leq v \leq 65$

$T(65,65)=(x=\frac{56}{65} (65)-\frac{33}{65} (65),\ \ y =\frac{33}{65} (65)+\frac{56}{65} (65)\\\\=(23,89)$

$T(-65,65)=(-56-33,\ \ -33+56)\\\\=(-89,23)$

$T(-65,-65) = (-56+33,-33-56)\\\\=(-23,-89)$

$T(65,-65)=(56+33, 33-56)\\\\=(89,-23)$

$\int \!\! \int_{T(S)} \ x^2 + y^2 \ {dA}$

$=\int\limits^{65}_{v=-65} \int\limits^{65}_{u=-65}(x^2+y^2)(\frac{d(x,y)}{d(u,v)} du\ \ dv$

Now

$x^2+y^2=(\frac{56}{65} u-\frac{33}{65} v)^2+(\frac{33}{65} u+\frac{56}{65} v)^2$

$[(\frac{56}{65} )^2+(\frac{33}{65}) ^2]u^2+[(\frac{33}{65} )^2+(\frac{56}{65}) ^2]v^2$

$=\frac{(65)^2}{(65)^2} u^2+\frac{(65)^2}{(65)^2} v^2=u^2+v^2$

$\int \!\! \int_{T(S)} \ x^2 + y^2 \ {dA}$

$=\int\limits^{65}_{v=-65} \int\limits^{65}_{u=-65}(u^2+v^2) du\ \ dv$

$=\int\limits^{65}_{-65}\int\limits^{65}_{-65}u^2du \ \ dv+\int\limits^{65}_{-65}\int\limits^{65}_{-65}v^2du \ \ dv$

By symmetry of the region

$=4\int\limits^{65}_0 \int\limits^{65}_0u^2 du \ \ dv + u\int\limits^{65}_0 \int\limits^{65}_0v^2 du \ \ dv$

$= 4(\frac{u^3}{3} )^{65}_{0}(v)_0^{65}+(\frac{v^3}{3} )^{65}_{0}(u)_0^{65}\\\\=4[\frac{(65)^4}{3} +\frac{(65)^4}{3} ]$

$=\frac{8}{3} (65)^4$

8 0

3 years ago