Solve for the base of a triangle where the height is 16cm and area is 104 cm2. What is the base?

Mathematics

2 answers:

Karo-lina-s [1.5K]2 years ago

7 0

Answer:

13 cm

Step-by-step explanation:

Area = 1/2×base×height

Formula for area:

S = 1.2bh

Solving for b:

b = 2s/h
b = 2*104/16 = 13 cm

Answer is 13 cm

Arturiano [62]2 years ago

6 0

Answer:

13cm

Step-by-step explanation:

104 × 2 = 208

208 ÷ 16 = 13cm

(13 × 16) ÷ 2 = 104cm²

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The length of a rectangular plot of land is 10 yards more than its width. If the area of the land 600 square yards , find the di

vagabundo [1.1K]

Let the width of the rectangular plot of land be 'x' yards. Given that the length of the rectangular plot of land is 10 yards more than its width. So, width of the rectangular plot of land = (x + 10) yards. Also given that the area of the rectangular plot of land is 600 square yards. We know that, area of a rectangle = length * width That is, (x+10) * x = 600 x^2 + 10x = 600 x^2 + 10x - 600 = 0 x^2 + 30x - 20x -600 = 0 x(x + 30) - 20(x + 30) = 0 (x +30)(x -20) =0 Therefore, either (x + 30) = 0 or (x - 20) = 0 If x + 30 = 0, then x = -30 and If x - 20 = 0, then x = 20 Since 'x' represents the width of a rectangular plot of land it cannot be negative. Therefore, width of the rectangular plot of land = 20 yards length of the rectangular plot of land= x + 10 = 30 yards

5 0

3 years ago

Find integra of xlnxdx

Mademuasel [1]

Answer:

$\dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

$\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\end{minipage}}$

Increase the power by 1, then divide by the new power.

Given indefinite integral:

$\displaystyle \int x \ln x \:\: \text{d}x$

To integrate the given integral, use Integration by Parts:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

$\text{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}$

$\text{Let }\dfrac{\text{d}v}{\text{d}x}=x \implies v=\dfrac{1}{2}x^2$

Therefore:

$\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x \ln x\:\:\text{d}x & = \ln x \cdot \dfrac{1}{2}x^2-\int \dfrac{1}{2}x^2 \cdot \dfrac{1}{x}\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x -\int \dfrac{1}{2}x\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}\end{aligned}$