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julia-pushkina [17]
4 years ago
8

Evaluate the expression 5t + 2m for m = 7 and t = 2.

Mathematics
1 answer:
pantera1 [17]4 years ago
5 0

Answer:

<h3>The answer is 24</h3>

Step-by-step explanation:

5t + 2m

m = 7 t = 2

Substitute the above values into the expression

That's

5(2) + 2(7)

10 + 14

24

Hope this helps you

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I need help, lmk if you can do them..
OLEGan [10]

Answer:

25) cos X = 12/37            cos X = 0.3243                         X = 71.08 degrees

26) tan C = 32/24            tan C = 1.3333                          C = 53.13 degrees

27) sin Z = 15/25              sin Z = 0.6                                   Z = 36.87 degrees

28) sin A = 32/40             sin A = 0.8                                 A = 53.13 degrees

29) sin Z = 40/50             sin Z = 0.8                                  Z = 53.13 degrees

7 0
3 years ago
Read 2 more answers
Solve triangle ABC. (If an answer does not exist, enter DNE. Round your answers to one decimal place. Below, enter your answers
OleMash [197]

Answer:

∠A1 = 27.4°, ∠A2 = 56.6°, ∠C1 =104.6°, ∠C2=75.4°, a1 = 79.9 and a2 = 144.9

Step-by-step explanation:

From Sine rule

\frac{a}{sinA}=\frac{b}{sinB} = \frac{c}{sinC}

∴ b / sinB = c / sinC

From the question,

b = 129, c = 168 and ∠B = 48°

∴ 129 / sin48° = 168 / sinC

Then, sinC = (168×sin48)/129

sinC = 0.9678

C = sin⁻¹(0.9678)

C = 75.42

∠C2=75.4°

and

∴∠C1 = 180° - 75.4°

∠C1 =104.6°

For ∠A

∠A1 = 180° - (104.6°+48°) [sum of angles in a triangle]

∠A1 = 27.4°

and

∠A2 = 180° - (75.4° + 48°)

∠A2 = 180° - (123.4°)

∠A2 = 56.6°

For side a

a1/sinA1 = b/sinB

a1/ sin27.4° = 129/sin48

a1 = (129×sin27.4°)/sin48

a1 = 79.8845

a1 = 79.9

and

a2/sinA2 = b / sinB

a2/ sin56.6° = 129/sin48

a2 = (129×sin56.6°)/sin48

a2 = 144.9184

a2 = 144.9

Hence,

∠A1 = 27.4°, ∠A2 = 56.6°, ∠C1 =104.6°, ∠C2=75.4°, a1 = 79.9 and a2 = 144.9

4 0
3 years ago
HURRY I'LL PICK A BRAINLIEST PERSON!!!!
german
The two correct answers are A and E. You can immediately eliminate C and E because to bring over something that is multiplied you have to divide (fraction). And B isn't the answer because to isolate s you divide by πr which would bring <span>πr to the denominator, not the numerator.
Hope this </span>helps!

5 0
3 years ago
Read 2 more answers
What is the equation in point−slope form of the line passing through (1, 2 and (2, 5?
Nimfa-mama [501]
Use the point-slope formula, y-y1=m(x-x1) to find the equation of the line
The answer is y=3x-1. Hope this helps!
7 0
3 years ago
Please help me solve this problem ASAP
DiKsa [7]

\bold{\huge{\blue{\underline{ Solution }}}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • <u>The </u><u>right </u><u>angled </u><u>below </u><u>is </u><u>formed </u><u>by </u><u>3</u><u> </u><u>squares </u><u>A</u><u>, </u><u> </u><u>B </u><u>and </u><u>C</u>
  • <u>The </u><u>area </u><u>of </u><u>square </u><u>B</u><u> </u><u>has </u><u>an </u><u>area </u><u>of </u><u>1</u><u>4</u><u>4</u><u> </u><u>inches </u><u>²</u>
  • <u>The </u><u>area </u><u>of </u><u>square </u><u>C </u><u>has </u><u>an </u><u>of </u><u>1</u><u>6</u><u>9</u><u> </u><u>inches </u><u>²</u>

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>area </u><u>of </u><u>square </u><u>A</u><u>? </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u><u> </u></h3>

The right angled triangle is formed by 3 squares

<u>We </u><u>have</u><u>, </u>

  • Area of square B is 144 inches²
  • Area of square C is 169 inches²

<u>We </u><u>know </u><u>that</u><u>, </u>

\bold{ Area \: of \: square =  Side × Side }

Let the side of square B be x

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 144 =  x × x }

\sf{ 144 =  x² }

\sf{ x = √144}

\bold{\red{ x = 12\: inches }}

Thus, The dimension of square B is 12 inches

<h3><u>Now, </u></h3>

Area of square C = 169 inches

Let the side of square C be y

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 169 =  y × y }

\sf{ 169 =  y² }

\sf{ y = √169}

\bold{\green{ y = 13\: inches }}

Thus, The dimension of square C is 13 inches.

<h3><u>Now, </u></h3>

It is mentioned in the question that, the right angled triangle is formed by 3 squares

The dimensions of square be is x and y

Let the dimensions of square A be z

<h3><u>Therefore</u><u>, </u><u>By </u><u>using </u><u>Pythagoras </u><u>theorem</u><u>, </u></h3>

  • <u>The </u><u>sum </u><u>of </u><u>squares </u><u>of </u><u>base </u><u>and </u><u>perpendicular </u><u>height </u><u>equal </u><u>to </u><u>the </u><u>square </u><u>of </u><u>hypotenuse </u>

<u>That </u><u>is</u><u>, </u>

\bold{\pink{ (Perpendicular)² + (Base)² = (Hypotenuse)² }}

<u>Here</u><u>, </u>

  • Base = x = 12 inches
  • Perpendicular = z
  • Hypotenuse = y = 13 inches

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ (z)² + (x)² = (y)² }

\sf{ (z)² + (12)² = (169)² }

\sf{ (z)² + 144 = 169}

\sf{ (z)² = 169 - 144 }

\sf{ (z)² = 25}

\bold{\blue{ z = 5 }}

Thus, The dimensions of square A is 5 inches

<h3><u>Therefore</u><u>,</u></h3>

Area of square

\sf{ = Side × Side }

\sf{ = 5 × 5  }

\bold{\orange{ = 25\: inches }}

Hence, The area of square A is 25 inches.

6 0
3 years ago
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