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3 years ago

When I evaluate m=2 , h=1 then what would mh+3 would be?

Mathematics

1 answer:

Andrei [34K]3 years ago

7 0

Answer:5

Step-by-step explanation: since you know that m=2, and h=1, mh is the same as m × h, or 2×1. Which is equal to 2, you then add 3 to get your answer

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Simplify: 2⁶X5⁴X3¹²/2²x5²x3⁴

Sholpan [36]

Step-by-step explanation:

2⁶X5⁴X3¹²/2²x5²x3⁴

2⁴ × 5²×3⁸

4 0

2 years ago

Read 2 more answers

Bob's password consists of a non-negative single-digit number followed by a letter and another non-negative single-digit number

maw [93]

9/20 is the probability that Bob's password consists of an odd single-digit number followed by a letter and a positive single-digit number

It is given that bob's password consists of a non-negative single-digit number followed by a letter and another non-negative single-digit number (which could be the same as the first one).

P (odd single digit number ) = ( 5 odd single digits / 10 non-negative single digits) = (5/10) = 1/2

P (positive single digit number ) = (9 positive single digits / 10 non-negative single digits) = 9/10

The total probability is (1/2) (9/10) = 9/20

Learn more about word problems here: brainly.com/question/21405634

#SPJ4

5 0

2 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

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

If cosA+sinA=√2, then prove A=45°.

Lyrx [107]

Answer:

1/2 now substitute 45 degrees

Step-by-step explanation:

6 0

2 years ago

Y/8 = -7 ? What is y ?

Nana76 [90]

Y = -56 i believe :)

8 0

3 years ago