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anzhelika [568]

3 years ago

8

if bill puts $3,500 into an account that has a 2.5% interest rate compounded annually, how much money will bill have after 3 yea

rs

1 answer:

lyudmila [28]3 years ago

8 0

Answer:

$26,250

Step-by-step explanation:

$3,500×(2.5)^3

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2. Find mzR and mzS.<br> 89°<br> R<br> 157 T<br> Р

baherus [9]

Answer:

m∠R = 123°

m∠S = 91°

Step-by-step explanation:

The measure of an inscribed angle is 1/2 the measure of the intercepted arc

m of arc QRS = 2 m∠P = 2(57) = 114

Now m of arc SPQ = 360 - 114 = 246

So, m∠R = 1/2(246) = 123

m of arc RSP = 2 m∠Q = 2(89) = 178

m of arc PQR = 360 - 178 = 182

m∠S = 1/2(182) = 91

5 0

3 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago

How can you solve this 60-20=20 ?

Stels [109]

Simplify each side in order to determine if true.

the answer is: False

5 0

3 years ago

Read 2 more answers

2/3 of a number -11. Please help

vfiekz [6]

Answer:

(2/3)x = -11

x = -33/2

Step-by-step explanation:

2/3 of a number is -11.

Let the unknown number be x.

2/3 of x is -11.

(2/3)x = -11

Multiply both sides by 3/2.

(3/2)(2/3)x = (3/2)(-11/1)

x = -33/2

8 0

3 years ago

Suppose △LMN≅△PQR .

Tanya [424]

First four answers are correct as congruent parts of congruent triangles are congruent but L Is not the opposite to M so they wouldn’t be congruent nor does NL or PQ have the same connection

6 0

3 years ago