State the place value of the underlined digit in 6 953 742. A Hundreds C Ten thousands B Thousands D Hundred thousands ( 3. Round off 5 987 341 to the nearest hundred thousand. A 5.8 million C 6.0 million ( B 5.9 million D 6.1 million 4. Which of the following numbers, when rounded off to the nearest thousand, becomes 7 541 000?
Solve the triangle. Given: A = 48° C = 97° a = 12 B = 35° b = 9.2 c = 16.0 5. The given measurements produce one triangle. Given: a = 7, b = 5, A = 70° C = 67.8, B = 42.1, c = 6.8 6. x×tan62 = (x+300)×tan53 = perpendicular x = 300×tan53/(tan62-tan53) = 719.0295 yards AB = (x+300/cos53 = 1019.0295/cos53 = 1693.3 yards Distance between A and B is 1693.3 yards 7. Given: sides, 5,6,7 in miles Island A = 78.46° Island B = 135.58° Island C = 57.12° Given the information… From Island B I would travel a Northwest Bearing to Island C. 8.
This equals -4/2, which reduces to -2. Now, we can use point-slope form and then rearrange it to make it into slope-intercept form. Since I’m using point C, it should look like this: y-(-4)=-2(x+1)=(y+4=-2x-2)=(y=-2x-6). So the answer in
a) 3x + 2y = 4 and x – 5y = 7 b) 3x – 5y = –11 and 5x + y = 19 3. Find the equation of the line parallel to 4x + y = 5 and passing through the point ( 8 , – 1 ) Find the equation of the line containing ( – 2 , 5 ) and perpendicular to the line containing the points ( 2 , 8 ) and ( 5 , – 1 ). Find an equation of the line perpendicular to the line 6x – 2y = – 5 and containing the point ( – 1 , – 2 ). 4. 5.
4 c. 1 d. 4 2. a. 5 b. 3 c. 4 d. 7 3. 2.63 x 10-6 4. a. 5 x 106 b.
x and c-x, then the diagram looks as follows: The perpendicular has divided the triangle into two right-angled triangles. Now for any right-angle triangle, according to Pythagorean Theorem, [pic] = [pic] + [pic] If Pythagorean is applied to the right-angled triangles in the above triangle, then in the case of left right-angle triangle in the above diagram, it would give us the equation [pic] = [pic] + [pic] where ‘a’ = hypotenuse and ‘h’ = height/perpendicular and ‘x’ = base. Re-writing it, the equation would become which we will call Eq. A [pic] = [pic] - [pic] ---------------------( Eq. A Similarly, for the right angle triangle on the right half to triangle ABC, [pic] = [pic] + [pic] where ‘b’ = hypotenuse, ‘h’ = height/perpendicular and ‘c-x’ = base.
Answer the next four questions using the following set of numbers. Show your work for all questions: 2.0, 3.7, 4.9, 5.0, 5.7, 6.7, 8.5, 9.0 What is the range of the above values? Range is the difference between the lowest value and the highest value. The range of the listed numbers is 9.0-2.0=7.0. What is the median of the above values?
Step 1) Identify the legs and the hypotenuse of the right triangle. | The legs have length '14' and 48 are the legs. The hypotenuse is X. See Picture | The hypotenuse is red in the diagram below: Steps 2 and 3 | Step 2) Substitute values into the formula (remember 'c' is the hypotenuse) | A2 + B2 = C2 142 + 482 = x2 | Step 3) Solve for the unknown | | Problem 2) Use the Pythagorean theorem to calculate the value of X. Round your answer to the nearest tenth.
5. If find f(g(x)). 6. Write the slope-intercept form of the equation of the line containing the points (4,7) and 7. Write the equation of the line perpendicular to , passing through the point (6,1).
One can check that v2 (16, 28) = 8, v2 (4, 16) = 1.25, v2 (4, 10) = 0.25 and v2 (1, 7) = 0. (iii) δ(s, y) = [vn+1 (2s, y + 2s) − vn+1 (s/2, y + s/2)]/[2s − s/2]. −→ 1.9: (i) Vn (w) = 1/(1+rn (w))[pn (w)Vn+1 (wH)+qn (w)Vn+1 (wT )] where pn (w) = (1 + rn (w) − dn (w))/(un (w) − dn (w)) and qn (w) = 1 − pn (w). (ii) ∆n (w) = [Vn+1 (wH) − Vn+1 (wT )]/[Sn+1 (wH) − Sn+1 (wT )] (iii) p = q = 1/2. V0 = 9.375.