Can anyone please help me? please?

Mathematics

1 answer:

ollegr [7]4 years ago

3 0

DHF is a larger angle making DF larger.

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Find the lengths of AB,BC,PQ,and QR. Which segments have equal lengths?

muminat

I need the picture that goes with this problem

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4 years ago

Rounded to the nearest hundredth , what is the positive solution to the quadratic equation 0=2x^2+3x-8

elena-14-01-66 [18.8K]

The positive solution to the quadratic equation $2 x^{2}+3 x-8=0$ is x = 1.39

<h3>Solution:</h3>

Given quadratic equation is $2 x^{2}+3 x-8=0$

The general quadratic equation is of form:

$a x^{2}+b x+c=0$

Now comparing the general equation with the given equation we get

a = 2 , b = 3 and c = -8

The formula to determine roots of the quadratic equation is:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

On plugging in vlaues, we get

$x=\frac{-3 \pm \sqrt{3^{2}-4 \times 2 \times(-8)}}{2 \times 2}$

$\mathrm{x}=\frac{-3 \pm \sqrt{9-(-64)}}{4}$

On solving we get,

$\begin{array}{l}{\mathrm{x}=\frac{-3 \pm \sqrt{9+64}}{4}} \\\\ {\mathrm{x}=\frac{-3 \pm \sqrt{73}}{4}}\end{array}$

$\mathrm{x}=\frac{-3+\sqrt{73}}{4} \text { OR } \mathrm{x}=\frac{-3-\sqrt{73}}{4}$

x = 1.39 OR x = -2.89

Hence , the positive solution to the quadratic equation is x = 1.39

7 0

4 years ago

T is the midpoint of SU. Find ST, TU and SU. 10x-14 5x+16 S———T———U

jeka94

since T is the midpoint of SU, then ST = TU.

$\bf \stackrel{10x-14}{\boxed{S}\rule[0.35em]{10em}{0.25pt}} T\stackrel{5x+16}{\rule[0.35em]{10em}{0.25pt}\boxed{U}} \\\\\\ \stackrel{ST}{10x-14}=\stackrel{TU}{5x+16}\implies 5x-14=16\implies 5x=30\implies x=\cfrac{30}{5}\implies x=6 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{ST}{10(6)-14\implies 46}~\hfill \stackrel{TU}{TU=ST=46}~\hfill \stackrel{SU}{ST+TU=92}$